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MA6459 NUMERICAL METHODS

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MA6459 NUMERICAL METHODS L T P C 3 1 0 4

OBJECTIVES:This course aims at providing the necessary basic concepts of a few numerical methods and give procedures for solving numerically different kinds of problems occurring in engineering and technologyUNIT I SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS 10+3Solution of algebraic and tranndental equations - Fixed point iteration method – Newton Raphson method- Solution of linear system of equations - Gauss elimination method – Pivoting - Gauss Jordan method – Iterative methods of Gauss Jacobi and Gauss Seidel - Matrix Inversion by Gauss Jordan method - Eigen values of a matrix by Power method.UNIT II INTERPOLATION AND APPROXIMATION 8+3Interpolation with unequal intervals - Lagrange's interpolation – Newton’s divided difference Interpolation – Cubic Splines - Interpolation with equal intervals - Newton’s forward and backward difference formulae.UNIT III NUMERICAL DIFFERENTIATION AND INTEGRATION 9+3Approximation of derivatives using interpolation polynomials - Numerical integration using Trapezoidal, Simpson’s 1/3 rule – Romberg’s method - Two point and three point Gaussian Quadrature formulae – Evaluation of double integrals by Trapezoidal and Simpson’s 1/3 rules. UNIT IV INITIAL VALUE PROBLEMS FOR ORDINARY DIFFERENTIALEQUATIONS

9+3Single Step methods - Taylor’s series method - Euler’s method - Modified Euler’s method – Fourth order Runge-Kutta method for solving first order equations - Multi step methods - Milne’s and Adams-Bash forth predictor corrector methods for solving first order equations.UNIT V BOUNDARY VALUE PROBLEMS IN ORDINARY AND PARTIAL

DIFFERENTIAL EQUATIONS 9+3Finite difference methods for solving two-point linear boundary value problems - Finite difference Techniques for the solution of two dimensional Laplace’s and Poisson’s equations on rectangular Domain – One dimensional heat flow equation by explicit and implicit (Crank Nicholson) methods– One dimensional wave equation by explicit method.TOTAL (L:45+T:15): 60 PERIODSOUTCOMES:The students will have a clear perception of the power of numerical techniques, ideas and would be able to demonstrate the applications of these techniques to problems drawn from industry, management and other engineering fields.TEXT BOOKS:1. Grewal. B.S., and Grewal. J.S.,"Numerical methods in Engineering and Science", Khanna Publishers, 9th Edition, New Delhi, 2007.2. Gerald. C. F., and Wheatley. P. O., "Applied Numerical Analysis", Pearson Education, Asia, 6th Edition, New Delhi, 2006.REFERENCES:

1. Chapra. S.C., and Canale.R.P., "Numerical Methods for Engineers, Tata McGraw Hill,5th Edition, New Delhi, 20072. Brian Bradie. "A friendly introduction to Numerical analysis", Pearson Education, Asia,New Delhi, 2007.3. Sankara Rao. K., "Numerical methods for Scientists and Engineers", Prentice Hall of IndiaPrivate, 3rd Edition, New Delhi, 2007.

Contents


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UNIT – I Solution of Equations & Eigen Value Problems 7

1.1 Numerical solution of Non-Linear Equations 7

Method of false position 8

Newton Raphson Method 9

Iteration Method 11

1.2 System of Linear Equations 13

Gauss Elimination Method 14

Gauss Jordan Method 15

Gauss Jacobi Method 17

Gauss Seidel Method 19

1.3 Matrix Inversion 20

Inversion by Gauss Jordan Method 21

1.4 Eigen Value of a Matrix 22

Von Mise`s power method 22

Tutorial Problems 24

Question Bank 26

UNIT- II Interpolation and approximation 30

2.1 Interpolation with Unequal Intervals 30

Lagrange`s Interpolation formula 31

Inverse Interpolation by Lagrange`s Interpolation Polynomial 32

2.2 Divided Differences-Newton Divided Difference Interpolation 34

Divided Differences 34

Newtons Divided Difference formula for unequal intervals 35

2.3 Interpolating with a cubic spline 37

Cubic spline interpolation 38

2.4 Interpolation with equals 40


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Newtons` forward interpolation formula 41

Newtons` Backward interpolation formula 41


Question Bank 46

UNIT- III Numerical Differentiation and Integration 50

3.1 Numerical Differentiation Derivatives using divided differences 50

Derivatives using finite Differences 51

Newton`s forward interpolation formula 51

Newton`s Backward interpolation formula 51

3.2 Numerical integration 53

Trapezoidal Rule 53

Simpson`s 1/3 Rule 54

Simpson`s 3/8 Rule 55

Romberg`s intergration 56

3.3 Gaussian quadrature 58

Two Point Gaussian formula & Three Point Gaussian formula 59

3.4 Double integrals 60

Trapezoidal Rule & Simpson`s Rule 61


Question Bank 66

UNIT – IV Numerical solution of Ordinary Differential Equation 67

4.1 Taylor`s Series Method 67

Power Series Solution 67

Pointwise Solution 68

Taylor series method for Simultaneous first order

Differential equations 70


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4.2 Euler Methods 72

Euler Method 73

Modified Euler Method 74

4.3 Runge – Kutta Method 74

Fourth order Runge-kutta Method 75

Runge-Kutta Method for second order differential equations 76

4.4 Multi-step Methods 78

Milne`s Method 79

Adam`s Method 80


UNIT – V Boundary Value Problems in ODE & PDE 87

5.1 Solution of Boundary Value Problems in ODE 87

5.2 Solution of Laplace Equation and Poisson Equation 90

Solution of Laplace Equation – Leibmann`s iteration process 91

Solution of Poisson Equation 92

5.3 Solution of One Dimensional Heat Equation 95

Bender-Schmidt Method 96

Crank- Nicholson Method 97

5.4 Solution of One Dimensional Wave Equation 99


Question Bank 106

Previous year University Question paper 107


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af (b)bf (a)f (b)f (a)

CHAPTER -1

SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS

1.1 Numerical solution of Non-Linear equations

IntroductionThe problem of solving the equation is of great importance in science and Engineering.

In this section, we deal with the various methods which give a solution for the equation

Solution of Algebraic and tranndental equationsThe equation of the form f(x) =0 are called algebraic equations if f(x) is purely a

polynomial in x.For example: are algebraic equations .If f(x) also contains trigonometric, logarithmic, exponential function etc. then the equation is known astranndental equation.

Methods for solving the equation

The following result helps us to locate the interval in which the roots ofMethod of false position. Iteration methodNewton-Raphson method

Method of False position (Or) Regula-Falsi method (Or) Linear interpolation method

In bisection method the interval is always divided into half. If a function changes sign over an interval, the function value at the mid-point is evaluated. In bisection method the interval from a to b into equal intervals ,no account is taken of the magnitude of .An alternative method that exploits this graphical insights is to join by a straight line.The intersection of this line with the X-axis represents an improved estimate of the root.The replacement of the curve by a straight line gives a “false position” of the root is the origin of the name ,method of false position ,or in Latin ,Regula falsi .It is also called the linear interpolation method.

Problems

1. Find a real root of that lies between 2 and 3 by the method of false position and correct to three decimal places.

Sol:Let

The root lies between 2.5 and 3.The approximations are given by

x

Iteration(r) a b xr f(xr)


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af (b)bf (a)f (b)f (a)

1 2.5 3 2.9273 -0.26642 2.927 3 2.9423 -0.00883 2.942 3 2.9425 -0.00544 2.942 3 2.9425 0.003

2. Obtain the real root of ,correct to four decimal places using the method of false position.

Sol:

GivenTaking logarithmic on both sides,

x log10 x 2f (x) x log10 x 2

f (3.5) 0.0958 0 and

f (3.6) 0.00027 0

The roots lies between 3.5 and 3.6

The approximation are given by x

Iteration(r) a b xr f(xr)1 3.5 3.6 3.5973 0.000015

2 3.5 3.5973 3.5973 0

The required root is 3.5973

Exercise:

1. Determine the real root of correct to four decimal places by Regula-Falsi method.Ans: 1.0499

2. Find the positive real root of correct to four decimals by the method of False position .Ans: 1.8955

3.Solve the equation by Regula-Falsi method,correct to 4 decimal places.Ans: 2.7984


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Newton’s method (or) Newton-Raphson method (Or) Method of tangents


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f (xi )f (xi )

f (xi )f (xi )xi 1 x

f (x0 )f (x0 )

x1x

1.1255.25

This method starts with an initial approximation to the root of an equation, a better and closer approximation to the root can be found by using an iterative process.

Derivation of Newton-Raphson formulaLet be the root of

f (x) 0 and x0 be an approximation to .If h x0

Then by the Tayor’s series

xi 1 xi ,i=0,1,2,3,………

Note:

The error at any stage is proportional to the square of the error in the previous stage.The order of convergence of the Newton-Raphson method is at least 2 or the convergence of N.R method is Quadratic.

Problems

1.Using Newton’s method ,find the root between 0 and 1 of correct five decimal places.

Sol:Given

f (x) x3 6x 4f (x) 3x 2 6

By Newton-Raphson formula, we have approximation

i , n 0,1,2,3.....

The initial approximation is x0 0.5

First approximation:

0 ,

0.5 ,


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=0.71429Second approximation:

f (x1 )f (x1 )

x2x

0.07874.4694

f (x2 )f (x2 )

x3x

0.00064.3923


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f (x3 )f (x3 )

x4x

0.0000034.3923

s

1 c

xi 1 xf (xi )f (xi )

sin xn0.5xn cos xn

1cos xn

1 ,

0.71429 ,

=0.7319

Third approximation

2 ,

0.73205 ,

Fourth approximation=0.73205

3 ,

0.73205 ,

=0.73205The root is 0.73205, correct to five decimal places.

2. Find a real root of x=1/2+sinx near 1.5, correct to 3 decimal places by newton-Rapson method. Sol:

f (x) x in x1

Let 2

f (x) osxBy Newton-Raphson formula, we have approximation

i

xn 1

, n 0,1,2,3.....

, n 0,1,2,3,....



Let x0 1.5 be the initial approximation

First approximation:

(1.5sin(1.5)0.5)1 cos1.51



1

c

1

x1 .5

1.5(0.0025 )(0.9293 )

=1.497

Second approximation:

x1 .497 ( 0.00028)0.9263

=1.497.

The required root is 1.497

Exercise:

1. Find the real root of ex 3xplaces.Ans:1.5121

,that lies between 1 and 2 by Newton’s method ,correct to 4 decimal

2.Use Newton-Raphson method to solve the equation 3xAns:0.6071

osx 1 0

3.Find the double root of the equation x3 x2

Ans: 0.03226x 1 0 by Newton’s method.

Iteration method (Or) Method of successive approximations(Or)Fixed point method

For solving the equation f (x) 0 by iteration method, we start with an approximationvalue of the root.The equation f (x) 0 is expressed as x (x) .The equation x (x) is called fixedpoint equation.The iteration formula is given by formula.

Theorem(Fixed point theorem)

xn 1 (xn ), n

0,1,2,..... called fixed point iteration

Let f (x) 0 be the given equation whose exact root is . The equation

f (x) 0 be rewritten as

x (x) .Let I be the interval containing the root x . If

(x) for all x in I, then the sequence of



Approximation x1 , x2 , x3 ,..... xn

will converges to . ,If the initial starting value

x0 is chosen in I.



11x

12(1x) 3

2

(

( 12(1x) 3

2

(

The order of convergence

TheoremLet . be a root of the equation x g(x)

.Ifg ( ) 0, g ( ) 0,....... g ( p 1) ( ) 0andg( p) ( ) 0

.then the convergence of iteration xi 1 g(xi ) is of order p.

NoteThe order of convergence in general is linear (i.e) =1

Problems

1.Solve the equation places.

Sol:

x3

x2 1 0 or the positive root by iteration method ,correct to four decimal

f (x) x3 x 2 1

f (0) 1 0andf (1) 1 0The root lies between 0 and 1.

x3 x 2 1

x

0 x 2 (1 x) 11

1 x

The given equation can be expressed as (x)

x)

x)

(0)

x)

1 and

21

(1)1

14 2

32(1 x) 2



( x)1 x(0,1)Choosing x0 0.75 ,the successive approximations are

x



11

1 0.75=0.75593

Hence the root is 0.7549

Exercise:

1. Find the cube root of 15, correct to four decimal places, by iteration method

Ans: 2.4662

1.2 System of linear equation

Introduction

Many problems in Engineering and science needs the solution of a system of simultaneous linearequations .The solution of a system of simultaneous linear equations is obtained by the following two types of methods

Direct methods (Gauss elimination and Gauss Jordan method)Indirect methods or iterative methods (Gauss Jacobi and Gauss Seidel method)

(a) Direct methods are those in which The computation can be completed in a finite number of steps resulting in the exact

solutionThe amount of computation involved can be specified in advance. The method is independent of the accuracy desired.

(b) Iterative methods (self correcting methods) are which Begin with an approximate solution andObtain an improved solution with each step of iterationBut would require an infinite number of steps to obtain an exact solution without round-off errorsThe accuracy of the solution depends on the number of iterations performed.

x2 0.75465x3 0.75463x4 0.75487x5 0.75488x6 0.75488



n n

Simultaneous linear equationsThe system of equations in n unknowns x1 , x2 , x3 ,.....

xn

is given by

a11 x1 a12 x2 .......... a1n xn b1

a21 x1 a22 x2 .......... a2n xn b2

…………………………………………………………

………………………………………………………….

………………………………………………………….

an1 x1 an2 x2 .......... amn xn bn

This can be written as AX B where A a ; X x , x , x ,.....x T ; B b ,b ,b ,....b. T

ij 1 2 3 n 1 2 3 n

This system of equations can be solved by using determinants (Cramer’s rule) or by means of matrices. These involve tedious calculations. These are other methods to solve such equations .In this chapter we will discuss four methods viz.

(i) Gauss –Elimination method(ii) Gauss –Jordan method(iii) Gauss-Jacobi method(iv) Gauss seidel method

Gauss-Elimination method

This is an Elimination method and it reduces the given system of equation to an equivalent upper triangular system which can be solved by Back substitution.Consider the system of equations

a11

x1

a21

x1

a31

x1

a12 x2

a22 x2

a32 x2

a13 x3 b1

a23 x3 b2

a33 x3 b3



Gauss-algorithm is explained below:

Step 1. Elimination of x1 from the second and third equations .If a11 0, the first equation isused to eliminate x1 from the second and third equation. After elimination, the reduced system is



Gauss e lim ination (U

a11 x1

a22 x2

a32 x2

a12 x2

a23 x3

a33 x3

a13 x3

b1 b2

b3

Step 2:Elimination of x2 from the third equation. If a22 0, We eliminate x2 from third equation and

the reduced upper triangular system is

a11 x1

a22 x2

a33 x3

a12 x2

a23 x3

b3

a13 x3

b1 b2

Step 3:From third equation x3 is known. Using x3 in the second equation x2 is obtained. using both

x2 And x3 in the first equation, the value of x1 is obtained.Thus the elimination method, we start with the augmented matrix (A/B) of the given system and

transform it to (U/K) by eliminatory row operations. Finally the solution is obtained by back substitution process.

Principle

( A / B) / K )

2. Gauss –Jordan method

This method is a modification of Gauss-Elimination method. Here the elimination of unknowns is performed not only in the equations below but also in the equations above. The co-efficient matrix A of the system AX=B is reduced into a diagonal or a unit matrix and the solution is obtained directly without back substitution process.

( A / B) Gauss Jordan (D / K )or(I / K )

Examples

1.Solve the equations 2x y 4z 12,8x y 2z 20,4x 11y 33 by (1)Gauss –Elimination



z3method (2) Gauss –Jordan method

Sol:



4z2827z27

7

(i)Gauss-Elimination method

The given system is equivalent to AX B where

and

Principle. Reduce to

=

From this ,the equivalent upper triangular system of equations is

2x y 4z 12

y 1

z=1,y=2,x=3 By back substitution.

The solution is x=3,y=2,z=1

(ii)Gauss –Jordan method

Principle. Reduce to

3z31



13z

From this ,we have

y = 2 z=1

Exercise

1. Solve the system by Gauss –Elimination method 5xAns: x= 39.3345;y=16.793;z=18.966

y 2z 42, x

y

30,2x

y

z 5 .

2. Solve the equations x y 9,2x y 4z 3,3x 4y 5z 40 by (1) Gauss –Eliminationmethod (2) Gauss –Jordan methodAns: x=1; y=3; z=5

Iterative method

These methods are used to solve a special of linear equations in which each equation must possess one large coefficient and the large coefficient must be attached to a different unknown in that equation.Further in each equation, the absolute value of the large coefficient of the unknown is greater than the sum of the absolute values of the other coefficients of the other unknowns. Such type of simultaneous linear equations can be solved by the following iterative methods.



Gauss-Jacobi method Gauss seidel method



c1 cc

cb

1

c1 a

b1 a

c1

c

b

1

b

a

1 a

cb

c

a1 x a2 x a3 x

b1 y b2 y b3 y

z d1

2 z d2

3 z d3

i.e., the co-efficient matrix is diagonally dominant.

Solving the given system for x,y,z (whose diagonals are the largest values),we

have

x [d1a1

1 y 1 z]

y [d 2

b2

2 x 2 z]

z [d3c3

3 x 3 y]

Gauss-Jacobi method

If the rth iterates are x(r ) , y (r ) , z (r ) ,then the iteration scheme for this method is

x(r 1) (d1a1

(r )1 z (r ) )

y (r 1) (d 2

b2

(r )2 z (r ) )

z (r 1) (d3c3

(r )3 y (r ) )

The iteration is stopped when the values x, y, z start repeating with the desired degree of accuracy.

Gauss-Seidel method

This method is only a refinement of Gauss-Jacobi method .In this method ,once a new value for a unknown is found ,it is used immediately for computing the new values of the unknowns.

If the rth iterates are, then the iteration scheme for this method is

x (r 1) 1 (d

a1

(r )1 z (r ) )

y 1

x 2

x 3

1 y 1



1

b

a

1 a

y (r 1) (d 2

b2

(r 1)2 z ( r ) )

z (r 1) (d3c3

(r 1)3 y (r 1) )

Hence finding the values of the unknowns, we use the latest available values on the R.H.S The process of iteration is continued until the convergence is obtained with desired accuracy.

x

2

x 3



z

28x4 yz322x17 y4z35x3y10z24

Conditions for convergence

Gauss-seidel method will converge if in each equation of the given system ,the absolute values o the largest coefficient is greater than the absolute values of all the remaining coefficients

naii aij i

j 1, j 1

1,2,3......... n

This is the sufficient condition for convergence of both Gauss-Jacobi and Gauss-seidel iteration methods.

Rate of convergence

The rate of convergence of gauss-seidel method is roughly two times that of Gauss-Jacobi method. Further the convergences in Gauss-Seidel method is very fast in gauss-Jacobi .Since the current values of the unknowns are used immediately in each stage of iteration for getting the values of the unknowns.

Problems

1.Solve by Gauss-Jacobi method, the following system28x 4ySol:

32, x

3y 10z 24,2x 17y 4z 35

Rearranging the given system as

The coefficient matrix is is diagonally dominant.

Solving for x,y,z ,we have

We start with initial values =The successive iteration values are tabulated as follows

1z



2

Iteration X Y Z1 1.143 2.059 2.4002 0.934 1.360 1.5663 1.004 1.580 1.8874 0.983 1.508 1.8265 0.993 1.513 1.8496 0.993 1.507 1.8477 0.993 1.507 1.847

X=0.993;y=1.507;z=1.847

2.Solve by Gauss-Seidel method, the following system27x 6ySol:

85;6x 15y 2z 72, x

y 54z 10

The given system is diagonally dominant.

Solving for x,y,z, we get

We start with initial values =The successive iteration values are tabulated as follows

Iteration X Y Z1 3.148 3.541 1.9132 2.432 3.572 1.9263 2.426 3.573 1.9264 2.425 3.573 1.9265 2.425 3.573 1.926

X=2.425;y=3.573;z=1.926

Exercise:

1.Solve by Gauss-Seidel method, the following system10 x 2 y z 9;2x 30 y 2z 44, x 3y 10 z 22Ans : x 0.89; y 1.341; z



2.780



2.Solve by Gauss-Seidel method and Gauss-Jacobi method, the following system30 x 2 y 3z 75;2x 2 y 18

z30, x

17 y 2z 48

Ans : (i)x

1.321; y 1.522; z

3.541(ii)x 2.580; y 2.798; z

1.069

1.3 Matrix

inversion Introduction

A square matrix whose determinant value is not zero is called a non-singular matrix. Every non- singular square matrix has an inverse matrix. In this chapter we shall find the inverse of the non-singular square matrix A of order three. If X is the inverse of A,Then

Inversion by Gauss-Jordan method

By Gauss Jordan method, the inverse matrix X is obtained by the following steps:

Step 1: First consider the augmented matrix Step 2: Reduce the matrix A in to the identity matrix I by employing row transformations.

The row transformations used in step 2 transform I to A-1

Finally write the inverse matrix A-1.so the principle involved for finding A-1 is as shown below

Note

The answer can be checked using the result

Problems

1.Find the inverse of by Gauss-Jordan method.

Sol:



X

Thus

Hence

2.Find the inverse of by Gauss-Jordan method

Exercise:

1. Find the inverse of by Gauss-Jordan method.

2.By Gauss-Jordan method ,find

A 1 if

1.4 Eigen value of a Matrix

Introduction

For every square matrix A, there is a scalar and a non-zero column vector X such thatAX .Then the scalar is called an Eigen value of A and X, the corresponding Eigen vector. We have studied earlier the computation of Eigen values and the Eigen vectors by means of analytical method. In this chapter, we will discuss an iterative method to determine the largest Eigen value and the corresponding Eigen vector.

Power method (Von Mise’s power method)

Power method is used to determine numerically largest Eigen value and the corresponding Eigen vector of a matrix A



Working Procedure

Yk 1 i

Xk i



(For a square matrix)

Assume the initial vector Then find Normalize the vector to get a new vector

(i.e) is the largest component in magnitude of Repeat steps (2) and (3) till convergence is achieved. The convergence of mi and Xi will give the dominant Eigen value and the

corresponding Eigen vector X Thus 1 lim ,i 0,1,2.......n.

And is the required Eigen vector.

Problems

1.Use the power method to find the dominant value and the corresponding Eigen vector

of the matrix

Sol:

Let

1 X 1

2 X 2

3 X 3



4 X 4

0



3

x

5 X 5

Hence the dominant Eigen value is 15.97 and the corresponding Eigen vector is

2. Determine the dominant Eigen value and the corresponding Eigen vector of

Using power method.

Exercise:

1. Find the dominant Eigen value and the corresponding Eigen vector of A Find

also the other two Eigen values.

2. Find the dominant Eigen value of the corresponding Eigen vector

of By power method. Hence find the other Eigen value also.

1. Determine the real root of Ans:1.0499

xex

Tutorial problemsTutorial 1

correct tot four decimal places by Regula falsi method.

2. Solve the equation xtanx=-1 by Regula falsi method ,correct to 4 decimal places Ans:2.7984

3. Find by Newton’s method , the real root of Ans:2.7406

x log10 x 1.2 correct to 4 decimal places

4. Find the double root of the equation x3

Ans:x=1

2 x 1 0 by Newton’s method.

5. Solve x 1 tan 1 x by iteration method ,starting with x 2Ans:2.1323

Tutorial 21.Solve the following system by Gauss-elimination method



x64xxx6xxxx

3z8

32

3

9x4x

5x1 x23 4 4; x1

7x2 3 4 12; x1 2 x3 4 5; x1 2 3 x4

Ans: x1

1, x2

2, x3

1, x4 2

2.Solve the system of equation x 2ymethod Ans:x=1,y=2,z=3.

,2x y 4z 20,4x y 2z 12 by Gauss-Jordan

3.Solve the following equations 10x 2y zBy Gauss-Seidel method.

9,2x 30y

z

44, 2x y 10z 22

4.Solve by Gauss Jacobi method ,the following28x 4y z 32,

xy 10z 24,2x y 4z 35

Ans:x=0.993,y=1.07,z=1.847.5.Solve the following equations13x1 5x2 3x3 4 18,2x1 12

x2

3x3 x4 30,3x1 4x2 10 x3

4 29,2x1 x2 3x3 x4 31

By Gauss-Seidel method.

Ans: x1

0.58, x2 3.482,x3 3.703, x4 4.163

Tutorial 3

1.Find the inverse of using Gauss-Jordan method.

Ans:

2.By Gauss-Jordan method ,find A-1 if

Ans:

3.Determine the dominant Eigen value and the corresponding Eigen vector of

using power method.Ans:[0.062,1,0.062]T

4.Find the smallest Eigen value and the corresponding Eigen vector



of using power method.Ans:[0.707,1,0.707]T

5.Find the dominant Eigen value and the corresponding Eigen vector of by power method.Hence find the other Eigen value also .Ans:λ=2.381



QUESTION BANK

PART A

1. What is the order of convergence of Newton-Raphson methods if the multiplicity of the root is one.Sol:

Order of convergence of N.R method is 2

2. Derive Newton’s algorithm for finding the pth root of a number N.Sol:

If x= N 1/p ,Then xp-N = 0 is the equation to be solved. Let f(x) = xp-N, f’(x) = px p-1

By N.R rule, if xr is the r th iterateXr+1 = xr -

= xr -

=

=

3. What is the rate of convergence in N.R method?Sol:

The rate of convergence in N.R method is of order 2

4. Define round off error.Sol:

The round off error is the quantity R which must be added to the finite representation of a computed number in order to make it the true representation of that number.

5. State the principle used in Gauss-Jordan method.Sol:

Coefficient matrix is transformed into diagonal matrix.6. Compare Gaussian elimination method and Gauss- Jordan method.Sol:

Gaussian elimination method Gauss- Jordan method1

23

Coefficient matrix is transformed into upper triangular matrixDirect methodWe obtain the solution by back substitution method

Coefficient matrix is transformed into diagonal matrixDirect methodNo need of back substitution method



7. Determine the largest eigen value and the corresponding eigen value vector of the matrix correct to two decimal places using power method.

Sol:

AX1 = = = 2 = 2X2

AX2 = = = 2 = 2X3

This shows that the largest eigen value = 2 The corresponding eigen value =

8. Write the Descartes rule of signsSol:

1) An equation f (x) = 0 cannot have more number of positive roots than there are changes of sign in the terms of the polynomial f (x) .2)An equation f (x) = 0 cannot have more number of positive roots than there are changes of sign in the terms of the polynomial f (x) .

9. Write a sufficient condition for Gauss seidel method to converge .(or) State a sufficient condition for Gauss Jacobi method to converge.

Sol:The process of iteration by Gauss seidel method will converge if in each equation of the

system the absolute value of the largest coefficient is greater than the sum of theabsolute values of the remaining coefficients.

10. State the order of convergence and convergence condition for NR method?Sol:The order of convergence is 2 Condition of convergence is

11. Compare Gauss Seidel and Gauss elimination method?Sol:

Gauss Jacobi method Gauss seidel method1.

2.3.

Convergence method is slow

Direct method Condition for convergence is the coefficient matrix diagonally dominant

The rate of convergence of Gauss Seidel method is roughly twice that of Gauss Jacobi.Indirect methodCondition for convergence is the coefficient matrix diagonally dominant

12) Is the iteration method a self correcting method always?Sol:

In general iteration is a self correcting method since the round off error is smaller.



13) If g(x) is continuous in [a , b] then under what condition the iterative method x = g(x) has a unique solution in [a , b].Sol:

Let x = r be a root of x = g(x) .Let I = [a , b] be the given interval combining the point x = r. if g′(x) for all x in I, the sequence of approximation x0 , x 1,......x nwill converge to the root r, provided that the initial approximation x0 is chosen in r.

14) When would we not use N-R method .Sol:

If x1 is the exact root and x0 is its approximate value of the equationf (x) = 0.we know that x 1= x0 –

If is small, the error will be large and the computation of the root by this, method will be a slow process or may even be impossible.Hence the method should not be used in cases where the graph of the function when it crosses the x axis is nearly horizontal.

15) Write the iterative formula of NR method.Sol:

Xn+1 = xn –

Part B

1. Using Gauss Jordan method, find the inverse of the matrix

2.Apply Gauss-seidel method to solve the equations 20x+y-2z=17: 3x+20y-z=-18: 2x-3y+20z=25.

3.Find a positive root of 3x- log 10 X =6, using fixed point iteration method.

4.Determine the largest eigen value and the corresponding eigen vector of the

matrix

with (1 0 0)T as the initial vector by power method.

5.Find the smallest positive root of the equation x = sin x correct to 3 decimal places using Newton-Raphson method.

6.Find all the eigen value and eigen vectors of the matrix using Jacobi method.

7.Solve by Gauss-seidel iterative procedure the system 8x-3y-2z=20: 6x+3y+12z=35: 4x+11y- z=33.



8.Find the largest eigen value of by using Power method.

9.Find a real root of the equation x3+x2-1=0 by iteration method.

10.Using Newton’s method, find the real root of x log 10 X=1.2 correct to five decimal places.

11.Apply Gauss elimination method to find the solution of the following system :2x+3y-z=5: 4x+4y-3z=3: 2x-3y+2z=2.

12. Find an iterative formula to find , where N is a positive number and hence find

13.Solve the following system of equations by Gauss-Jacobi method: 27x+6y-z=85:x+y+54z=110: 6x+15y+2z=35.

14.Find the Newton’s iterative formula to calculate the reciprocal of N and hence find the value of.

15.Apply Gauss-Jordan method to find the solution of the following system: 10x+y+z=12: 2x+10y+z=13: x+y+5z=7.



CHAPTER 2

INTERPOLATION AND APPROXIMATION

2.1 Interpolation with Unequal

intervals Introduction

Interpolation is a process of estimating the value of a function at ana intermediate point when its value are known only at certain specified points.It is based on the following assumptions:

(i) Given equation is a polynomial or it can be represented by a polynomial with a good degree of approximation.

(ii) Function should vary in such a way that either it its increasing or decreasing in the given range without sudden jumps or falls of functional values in the given interval.

We shall discuss the concept of interpolation from a set of tabulated values of when the values of x are given intervals or at unequal intervals.First we consider interpolation with unequal intervals.

Lagrange’s Interpolation formula

This is called the Lagrange’s formula for Interpolation.

Problems

1. Using Lagrange’s interpolation formula, find the value of y corresponding to x=10 from the following data

X 5 6 9 11Y 12 13 14 16

Sol:

Given



By Lagrange’s interpolation formula

Using x=10 and the given data,

y (10) =14.67

2. Using Lagrange’s interpolation formula, find the value of y corresponding to x=6 from the following data

X 3 7 9 10Y 168 120 72 63

Sol:

Given


Using x=6 and the given data, y(6)=147

3.Apply Lagrange’s formula to find f(5),given that f(1)=2,f(2)=4,f(3)=8 and f(7)=128

Sol:

Given




Using x=5 and the given data, y(5)=32.93

Exercise

1.Find the polynomial degree 3 fitting the following data

X -1 0 2 3Y -2 -1 1 4

Ans:

2.Given

Ans:

Inverse Interpolation by Lagrange’s interpolating polynomial

Lagrange’s interpolation formula can be used find a value of x corresponding to a given y which is not in the table. The process of finding such of x is called inverse interpolation.

If x is the dependent variable and y is the independent variable, we can write a formula for x as a function of y.

The Lagrange’s interpolation formula for inverse interpolation is

Problems

1.Apply Lagrange’s formula inversely to obtain the root of the

equation Given that

Sol:

Given that



To find x such that

Applying Lagrange’s interpolation formula inversely ,we get

=0.8225

2.Given data

x 3 5 7 9 11y 6 24 58 108 174Find the value of x corresponding to y=100.

Sol:

Given that

Using the given data and y=100, we get

The value of x corresponding to y=100 is 8.656X=8.6



2.2 Divided Difference –Newton Divided Difference Interpolation

Formula Introduction

If the values of x are given at unequal intervals, it is convenient to introduce the idea of divided differences. The divided difference are the differences of y=f(x) defined, taking into consideration the changes in the values of the argument .Using divided differences of the function y=f(x),we establish Newton’s divided difference interpolation formula, which is used for interpolation which the values of x are at unequal intervals and also for fitting an approximate curve for the given data.

Divided differenceLet the function assume the valuesCorresponding to the arguments x1 , x2 , x3 ,.....

xn

need to be equal.

respectively, where the intervals

DefinitionsThe first divided difference of f(x) for the arguments is defined by

It is also denoted by [ ]Similarly for arguments and so on.The second divided differences of f(x) for three arguments Is defined as

x0 , x1 , x2

The third divided difference of f(x) for the four arguments Is defined as

x0 , x1 , x2 , x3

And so on .



Representation by Divided difference table

Arguments Entry First Divided difference Second D-D Third D-D

Properties of divided difference

The divided differences are symmetrical in all their arguments. The operator is linear. The nth divided differences of a polynomial of nth degree are constants.

Problems

1.If find the divided difference

Sol:

=



=

=

2.Find the third differences with arguments 2,4,9,10 for the function

Newton’s Divided difference Formula (Or) Newton’s Interpolation Formula for unequal intervals

Problems

1.Find the polynomial equation passing through(-1,3),(0,-6),(3,39),(6,822),(7,1611) Sol:Given data

x -1 0 3 6 7y 3 -6 39 822 1611The divided difference table is given as follows

-1

0

3

6

7

3

-6

39

822

1611

-9

15

261

789

6

41

132

5

131



By Newton’s formula,

Is the required polynomial.

2. Given the data

x 0 1 2 5f(x) 2 3 12 147

Find the form of the function .Hence find f(3).

Ans: f(3)=35

2.3 Interpolating with a cubic spline-cubic spline Interpolation

We consider the problem of interpolation between given data points (xi, yi),

i=0, 1, 2, 3….n where a= x1 x2 x3

.....

xn b by means of a smooth polynomial curve.

By means of method of least squares, we can fit a polynomial but it is appropriate.”Spline fitting” is the new technique recently developed to fit a smooth curve passing he given set of points.

Definition of Cubic spline

A cubic spline s(x) is defined by the following properties.

S(xi)=yi,i=0,1,2,…..n S(x),s’(x),s”(x) are continuous on [a,b] S(x) is a cubic polynomial in each sub-interval(xi,xi+1),i=0,1,2,3…..n=1

Conditions for fitting spline fit

The conditions for a cubic spline fit are that we pass a set of cubic through the points, using a new cubic in each interval. Further it is required both the slope and the curvature be



the same for the pair of cubic that join at each point.

)



6si 1

1

6 1s

Natural cubic spline

A cubic spline s(x) such that s(x) is linear in the intervals (i.e. s1=0 and sn=0 is called a natural cubic spline

, x1 )and(xn ,

where s1 =second derivative at (x1 , y1 )

Notesn Second derivative at (xn , yn )

The three alternative choices used are

s1 0 and sn 0 i.e. the end cubics approach linearity at their extremities.

s1 s2 ; sn sn 1 i.e. the end cubics approaches parabolas at their extremitiesTake s1 as a linear extrapolation from s2 and s3 andsn is a linear extrapolation from sn 1andsn 2

.with the assumption, for a set of data that are fit by a single cubic equation their cubic splines will all be this same cubic

For equal intervals ,we have hi 1

hi

h ,the equation becomes

si 1 4si 2 [ yi 1h

2 yi

yi 1 ] for i=1,2,3,….n-1

Problems

1.Fit a natural cubic spline to the following data

x 1 2 3y -8 -1 18And compute (i) y(1.5) (ii) y’(1)

Solution

Here n=2 ,the given data (x1 , y1 ) (1, 8), (x2 , y2

)(2, ), (x3 , y3

)(3,18)

For cubic natural spline, s1

0 & s3

0 .The intervals are equally spaced.

For equally spaced intervals,the relation on s1 , s2 & s3 is given by

s1 4s23

12[ y1 2

y2



s218

y3 ][

h ]

For the interval 1 x 2[x1 x 2],the cubic spline is given by

y a1 (x x1 ) b1 (x x1 ) c1 (x x1 ) d13 2



(s2s1

s1

8

y

h1 1( yi 1 yi )

(s1

The values of a1 ,b1 , c1 , d1

are given by

a1 ,b16

, c1

2

y2 y1

)(

2s16

s2),d1

y1

[ hi 1]

a1

3,b1

0, c1

4, d1

The cubic spline for the interval 1 x 2 is

s(x) 3(x 1)3 4(x 14) 8

(1.5) s(1.5)45

8

The first derivative of s(x) is given by

s (xi ) (2si6

si 1 )hi

Taking i=1, s (1) (2s16 2 ) y2 y1 ) [ hi 1]

s (1) 4

y (1) 4

We note that the tabulated function is y x3 9 and hence the actual values of y(1.5) and y (1)

are respectively45

and 3.8

2.The following values of x and y are given ,obtain the natural cubic spline which agree with y(x) at the set of data points

x 2 3 4y 11 49 123Hence compute (i) y(2.5) and (ii) y (2)

Exercise:



1.Fit the following data by a cubic spline curve

x 0 1 2 3 4y -8 -7 0 19 56Using the end condition that s1 & s5 are linear extrapolations.

y

y y 2 y y

2

y



yn 1y1

2

2

y0

f

2.Fit a natural cubic spline to

f (x) 201 5x

2

on the interval [-2,-1].Use five equispaced points of

the function at x=-2(1)2.hence find y(1.5).

2.4 Interpolation with equal intervals

(Newton’s Forward and Backward Difference formulas)

Introduction

If a function y=f(x) is not known explicitly the value of y can be obtained when s set ofvalues of xi , yi i=1,2,3….n are known by using methods based on the principles of finitedifferences ,provided the function y=f(x) is continuous.Assume that we have a table of values xi , yi i=1,2,3….n of any function, the values of x being

equally spaced, i.e., xi

x0 ih, i

0,1,2,....n

Forward differencesIf y0 , y1 , y2 ,...... yn denote a set of values of y,then the first forward differences of y=f(x)

are defined by y0

y1

y0 ; y2 y1 ;......... .

yn yn 1

Where is called the forward difference operator.

Backward differencesThe differences y1 y0 , y2 y1 ,……….. yn yn 1 are called first backward

differences and they are denoted by yn 2.9 y1

y1

y0 ; y2

y2

y1 ;......... .

yn

yn

yn 1

Where is called the backward difference operator.In similar way second, third and higher order backward differences are defined.

2 y2 y1 ; 3 y3 y2 ;......... . y4 y4 y3 .......3 2 2 3 2

3 3 4 4 y3 and so on.

Fundamental Finite Difference operators

Forward Difference operator ( y1 y0 )

If =f(x) then y f (x h) (x)Where h is the interval of differencing

y2

r



f

Backward Difference operator ( y1

y1

y0 )

The operator is defined by f (x) f (x) (x h)

Shift operatorThe shift operator is defined by Eyr yr 1

i.e the effect of E is to shift functional value yr

to the next value yr 1 .Also E 2 y y 2

y



1

1E 1

xx0 h

p y0p( p1)( p2) ..

p yn ..

xxn h

In general En y yn r

The relation between and E is given byE 1(or)E

Also the relation between and E-1 is given by

Newton’s Forward Interpolation formulaLet y=f(x) be a function which takes the values y0 , y1 , y2 ,...... yn corresponding to

the values x0 , x1 , x2 ,...... xn where the values of x are equally spaced.

i.e. xi x0 ih,i 0,1,2,......., n

y p y0

p

p( p 1) 2

2!0 . .......

3!

This formula is called Newton-Gregory forward interpolation formula.

Newton’s Backward Interpolation formulaThis formula is used for interpolating a value of y for given x near the end of a table

of values .Let y0 , y1 , y2 ,...... yn

be the values of y =f(x) for x x0 , x1 , x2 ,...... xn

where

xi x0

(ih,i)

0,1,2,......., np( p 1) 2 p( p 1)(

p2) 3 ...

y xn ph yn

p

yn yn2! 3!

This formula is called Newton Backward interpolation formula.We can also use this formula to extrapolate the values of y, a short distance ahead of yn.

r

y



y0 02 y

xx0 h

p0.2

..p y0

Problems

1. Using Newton’s Forward interpolation formula ,find f(1.02) from the following data

X 1.0 1.1 1.2 1.3 1.4F(x) 0.841 0.891 0.932 0.964 0.985Sol:

Forward difference table:

x y f (x) y 2 y 3 y

1.0

1.1

1.2

1.3

1.4

0.841

0.891

0.932

0.964

0.985

0.050

0.041

0.032

0.021

-0.009

-0.009

-0.011

0

-0.002

y0 0.841, 0.050 , 0 .009

Let x0 1.0 & x

p

1.02

By Newton’s Forward interpolation formula,

y(1.02) y0 p( p 1) 2

2! 0 ......

=0.852y(1.02) =0.852

2.Using Newton’s Forward interpolation formula, find the value of sin52 given thatsin 45 0

0.7071, sin 50 0

0.7660 , sin 55 0

0.8192 , sin 60 0

0.8660 . Ans:0.7880

3.Using Newton’s Backward interpolation formula, find y when x=27,from the following data

x 10 15 20 25 30



y 35.4 32.2 29.1 26.0 23.1

yn

2



2 4

..p yn

xxn h

p0.6

Sol:

Backward difference table

x y y 2 y 3 y 4 y

10

15

20

25

30

35.4

32.2

29.1

26.0

23.1

-3.2

-3.1

-3.1

-2.9

0.1

0

0.2

-0.1

0.20.3

Here xn 30, yn 23.1, yn 2.9 yn 0.2, 3 0.2 ,

yn 0.3

By Newton’s backward interpolation formula,

y(xn

ph) ynp( p 1)

y2! np( p 1)(

p3!

2) 3n ...

Here x=27, p

y(27)=24.8

4.Find the cubic polynomial which takes the value y(0)=1,y(1)=0,y(2)=1,y(3)=10.Hence or otherwise ,obtain y(4).Ans:y(4)=33

Exercise: 1.Given the datax 0 1 2 3 4y 2 3 12 35 78Find the cubic function of x,using Newton’s backward interpolation formula.

Ans: y(x) x3 x 2 x 2

2. Using Newton’s Gregory backward formula, find e 1.9 from the following data

y



x 1.00 1.25 1.50 1.75 2.00

e x 0.3679 0.2865 0.2231 0.1738 0.1353



Ans: e 1.9 =0.1496

3.Estimate exp(1.85) from the following table using Newton’s Forward interpolation formula

x 1.7 1.8 1.9 2.0 2.1 2.2 2.3

e x 5.474 6.050 6.686 7.839 8.166 9.025 9.974

Ans:6.3601

4.Find the polynomial which passes through the points (7,3)(8,1)(9,1)(10,9)using Newton’s interpolation formula.

Tutorial Problems

Tutorial 1

1.Apply Lagrange’s formula inversely to obtain the root of the equation f(x)=0,given that f(0)=-4,f(1)=1,f(3)=29,f(4)=52.Ans: 0.8225

2.Using Lagrange’s formula of interpolation ,find y(9.5) given the dataX 7 8 9 10Y 3 1 1 9Ans: 3.625

3.Use Lagrange’s formula to find the value of y at x=6 from the following datax 3 7 9 10y 168 120 72 63Ans: 147

4.If log(300)=2.4771,log(304)=2.4829,log(305)=2.4843,log(307)=2.4871find log(301)Ans: 2.8746

5. Given u0

Ans:96, u1

9, u3

33, u7

15..Find u 2



Tutorial 2

1.Given the datax 0 1 2 5F(x) 2 3 12 147Find the form of the function. Hence find f(3).Ans:35

2.Find f(x) as a polynomial in powers of(x-5) given the following tableX 0 2 3 4 7 9Y 4 26 58 112 466 922Ans: f (x) (x 5)3 17(x 5)2 98(x 5) 194

3.from the following table,find f(5)X 0 1 3 6F(x) 1 4 88 1309Ans:676

4.Fit the following data by a cubic spline curveX 0 1 2 3 4Y -8 -7 0 19 56Using the end condition that s1 and s5 are linear extrapolations. Ans:-8

5.Given the dataX 1 2 3 4Y 0.5 0.333 0.25 0.20Find y(2.5) using cubic spline function. Ans:0.2829

Tutorial 31. from the following table,find the number of students who obtained less than 45 marks

Marks 30-40 40-50 50-60 60-70 70-80No of students 31 42 51 35 31Ans:48

2.Find tan(0.26) from the following valuesof tanx for 0.10 x 0.30X 0.10 0.15 0.20 0.25 0.30Tanx 0.1003 0.1511 0.2027 0.2553 03093Ans:0.2662

3.Using newton’s interpolation formula find (i) y when x=48 (ii) y when x=84 from the following dataX 40 50 60 70 80 90Y 184 204 226 250 276 304Ans:199.84,286.96



4.Find the value of f(22) and f(42) from the following dataX 20 25 30 35 40 45F(x) 354 332 291 260 231 204Ans:352,219

5.Estimate sin 380 from the following data given belowx 0 10 20 30 40sinx 0 0.1736 0.3240 0.5000 0.6428Ans:12.7696

Question BankPart A

1. State the Lagrange’s interpolation formula.Sol:

Let y = f(x) be a function which takes the values y0, y1,……yn corresponding to x=x0,x1,……xn

Then Lagrange’s interpolation formula is

Y = f(x) =

yo

+ y1

+ ………+ yn

2. What is the assumption we make when Lagrange’s formula is used?Sol:

Lagrange’s interpolation formula can be used whether the values of x, the independent variable are equally spaced or not whether the difference of y become smaller or not.

3. When Newton’s backward interpolation formula is used.Sol:

The formula is used mainly to interpolate the values of y near the end of a set of tabular values and also for extrapolation the values of y a short distance ahead of y0

4. What are the errors in Trapezoidal rule of numerical integration?Sol:

The error in the Trapezoidal rule is



E< y”

5. Why Simpson’s one third rule is called a closed formula?



Sol:Since the end point ordinates y0 and yn are included in the Simpson’s 1/3 rule, it is called

closed formula.

6. What are the advantages of Lagrange’s formula over Newton’s formula?Sol:

The forward and backward interpolation formulae of Newton can be used only when the values of theindependent variable x are equally spaced and can also be used when the differences of the dependent variable y become smaller ultimately. But Lagrange’s interpolation formula can be used whether the values of x, the independent variable are equally spaced or not and whether the difference of y become smaller or not.

7. When do we apply Lagrange’s interpolation?Sol:

Lagrange’s interpolation formula can be used when the values of “x” are equally spaced or not. It is mainly used when the values are unevenly spaced.

8. When do we apply Lagrange’s interpolation?Sol:

Lagrange’s interpolation formula can be used when the values of “x” are equally spaced or not. It is mainly used when the values are unevenly spaced.

9. What are the disadvantages in practice in applying Lagrange’s interpolation formula?Sol:

1. It takes time.2. It is laborious

10. When Newton’s backward interpolation formula is used.Sol:

The formula is used mainly to interpolate the values of ‘y’ near the end of a set of tabularvalues.

11. When Newton’s forward interpolation formula is used.Sol:

The formula is used mainly to interpolate the values of ‘y’ near the beginnig of a set of tabular values.

12. When do we use Newton’s divided differences formula?Sol: This is used when the data are unequally spaced.

13. Write Forward difference operator.Sol:

Let y = f (x) be a function of x and let of the values of y. corresponding to of the values of x. Here, the independent variable (or argument), x

proceeds at equally spaced intervals and h (constant),the difference between two consecutive values of x is called the interval of differencing. Now the forward difference operator is defined as



==

......................=

These are called first differences.

14.Write Backward difference operator.Sol:

The backward difference operator is defined as=

For n=0,1,2 …===

………………….These are called first differences

Part B

1.Using Newton’s divided difference formula, find f(x) from the following data and hence find f(4).

x 0 1 2 5f(x) 2 3 12 147.

2.Find the cubic polynomial which takes the following values:

x 0 1 2 3f(x) 1 2 1 10

3.The following values of x and y are given:

x 1 2 3 4f(x) 1 2 5 11Find the cubic splines and evaluate y(1.5) and y’(3).

4.Find the rate of growth of the population in 1941 and 1971 from the table below.

Year X 1931 1941 1951 1961 1971Population Y

40.62 60.8 79.95 103.56 132.65

5.Derive Newton’s backward difference formula by using operator method.

6.Using Lagrange’s interpolation formula find a polynomial which passes the points (0,-12),(1,0),(3,6),(4,12).



7.Using Newton’s divided difference formula determine f(3) from the data:

x 0 1 2 4 5f(x) 1 14 15 5 68.Obtain the cubic spline approximation for the function y=f(x) from the following data, given that y0” = y3”=0.

x -1 0 1 2y -1 1 3 359.The following table gives the values of density of saturated water for various temperatures of saturated steam.

Find by interpolation, the density when the temperature is 2750 .

10.Use Lagrange’s method to find log 10 656 , given that log 10 654 =2.8156,log 10 658 =2.8182 , log 10 659 =2.8189 and log 10 661 =2.8202.

11.Find f’(x) at x=1.5 and x=4.0 from the following data using Newton’s formulae for differentiation.

x 1.5 2.0 2.5 3.0 3.5 4.0Y=f(x) 3.375 7.0 13.625 24.0 38.875 59.012.If f(0)=1,f(1)=2,f(2)=33 and f(3)=244. Find a cubic spline approximation, assuming M(0)=M(3)=0.Also find f(2.5).

13.Fit a set of 2 cubic splines to a half ellipse described by f(x)= [25-4x2]1/2. Choose the three data points (n=2) as (-2.5,0), (0,1.67) and (2.5 , 0) and use the free boundary conditions.

14.Find the value of y at x=21 and x=28 from the data given below

x 20 23 26 29y 0.3420 0.3907 0.4384 0.484815. The population of a town is as follows:

x year 1941 1951 1961 1971 1981 1991

y population (thousands)

20 24 29 36 46 51

Estimate the population increase during the period 1946 to1976.

Temperature 0

C100 150 200 250 300

Density hg/m3 958 917 865 799 712



CHAPTER 3

NUMERICAL DIFFERENTIATION AND INTEGRATION

Introduction

Numerical differentiation is the process of computing the value of

For some particular value of x from the given data (xi,yi), i=1, 2, 3…..n where y=f(x) is not known explicitly. The interpolation to be used depends on the particular value of x which derivatives are required. If the values of x are not equally spaced, we represent the function by Newton’s divided difference formula and the derivatives are obtained. If the values of x are equally spaced, the derivatives are calculated by using Newton’s Forward or backward interpolation formula. If the derivatives are required at a point near the beginning of the table, we use Newton’s Forward interpolation formula and if the derivatives are required at a point near the end of table. We use backward interpolation formula.

3.1 Derivatives using divided differences

Principle

First fit a polynomial for the given data using Newton’s divided difference interpolation formula and compute the derivatives for a given x.

Problems

1.Compute f (3.5)andf (4) given that f(0)=2,f(1)=3;f(2)=12 and f(5)=147

Sol:

Divided difference table

x f 2

1 f3

1 f0

1

2

5

2

3

12

147

1

9

45

4

91

Newton’s divided difference formula is

y

y

y

y



1

dydx

x x

1 2 y0 3 y0 4 y0 ..y0

..d 2 ydx2

x x

1

d 3 ydx3

x x

1 ..

dydx x x

1 2 yn 3 yn 4 yn ..yn

..d 2 ydx 2

x x

1

d 3 ydx3

x x

1 ..

f (x)f (x0 ) (x x0 ) f (x0 , x1 ) (x x0 )(

xx1 ) f (x0 , x1 , x2

).......

Here x0 0; x1

1,; x2 2, f (x0 ) 2, f (x0 , x1 ) =1, f (x0 , x1 , x2

)4; f (x0 , x1 , x2 , x3 ) 1

f (x) x3 x 2 x 2

f (x)f (x)

3x2 2x6x 2

f (3.5) 42.75, f (4) 26.

2.Find the values

f (5)andf (5) using the following data

x 0 2 3 4 7 9F(x) 4 26 58 112 466 22Ans: 98 and 34.

3.1.1 Derivatives using Finite differences

Newton’s Forward difference Formula

[ .....h 2

0

[ 2 yh 2 0

0

3 4

3 11 4

0 12 0......]

[ 3 yh3 0

0

3 4 y2 0

...... ........]

Newton’s Backward difference formula

[ .......].h 2

n

[ 2 y

h 2 nn

3 4

3 11 4

n 12 n...... ...]

[ 3 y

h3 nn

3 4 y2 n

...... ........]

y y



dydx

x x

1 2 yn 3 yn 4 yn ..yn

2

dydx

x 54

0

d 2 ydx2

x 54

1 ..

Problems

1.Find the first two derivatives of y at x=54 from the following data

x 50 51 52 53 54y 3.6840 3.7083 3.7325 3.7563 3.7798Sol:

Difference table

x y y 2 y 3 y 4 y

50

51

52

53

54

3.6840

3.7083

3.7325

3.7563

3.7798

0.0244

0.0241

0.0238

0.0235

-0.0003

-0.0003

-0.0003

0

00

By Newton’s Backward difference formula

[ .......].n

Here h=1;

h 2

yn =0.0235; yn

0.02335

3 4

.0003

[ 2 y

h2 n

3 11 4

n 12 n..........

=-0.0003



dy1 y0

pxxh

0

p0.2

dydx

x 1

1

d 2 ydx2

x 1

2.Find first and second derivatives of the function at the point x=12 from the following data

x 1 2 3 4 5y 0 1 5 6 8

Sol:

Difference table

x y y 2 y 3 y 4 y1

2

3

4

5

0

1

5

6

8

1

4

1

2

3

-3

1

-6

410

We shall use Newton’s forward formula to compute the derivatives since x=1.2 is at the beginning.

For non-tabular value x x0

ph ,we have

2 3 2

[2 p 1 2 y

3 p 6 p 2 3 y2 p 9 p 11 p 3 4 y .......

dx h 2 0 6 0 12 0

Here x0 =1, h=1, x=1.2

.773.2

14.17.2

2x



ny ..

h ....

Exercise:

1.Find the value of sec310 using the following data

x 31 32 33 34Tanx 0.6008 0.6249 0.6494 0.6745

Ans:1.1718

2.Find the minimum value of y from the following table using numerical differentiation

x 0.60 0.65 0.70 0.75y 0.6221 0.6155 0.6138 0.6170Ans: minimum y=0.6137

3.2 Numerical integration(Trapezoidal rule & Simpson’s rules, Romberg integration)

IntroductionThe process of computing the value of a definite integral from a set of values

(xi,yi),i=0,1,2,….. Where x0 =a; xn b of the function y=f(x) is called Numerical integration.Here the function y is replaced by an interpolation formula involving finite differences and then integrated between the limits a and b, the value is found.

General Quadrature formula for equidistant ordinates (Newton cote’s formula)

On simplification we obtainxn 2

ydx nh[ y

n(2n 3) 2 yn(n 2) 3 y .....

0 0 1200 24 0

This is the general Quadrature formulaBy putting n=1, Trapezoidal rule is obtained By putting n=2, Simpson’s 1/3 rule is derived By putting n=3, Simpson’s 3/8 rule is derived.

NoteThe error in Trapezoidal rule is of order h2 and the total error E is given by y”( ) is the largest of y.

Simpson’s 1/3 rulexn

ydxx0

[(y03

yn ) 4( y1 y3 y5 .... yn 1 ) 2 (y2 y4 y6



.... yn 2 )]

....3



I h 21 2 I 2 1

h 2

h 22

h 21

4I 2I1

Error in Simpson’s 1/3 ruleThe Truncation error in Simpson’s 13 rule is of order h4 and the total error is given byE

(b a) h4 yiv (x)whereyiv (x) is the largest of the fourth derivatives.

180

Simpson’s 3/8 rulexn

ydxx0

3h [(y

8 0 yn ) ( y1 y2 y4 ....) 2(y3 y6 y9 ....)

NoteThe Error of this formula is of order h5 and the dominant term in the error is given by

3 h5 yiv (x)

80

Romberg’s integrationA simple modification of the Trapezoidal rule can be used to find a better approximation to

the value of an integral. This is based on the fact that the truncation error of the Trapezoidal rule is nearly proportional to h2.

I

This value of I will be a better approximation than I1 or I2.This method is called Richardson’s deferred approach to the limit.If h1=h and h2=h/2,then we get

I (h, h

)2 3

Problems

1.Evaluate sin xdx by dividing the interval into 8 strips using (i) Trapezoidal rule0

(ii) Simpson’s 1/3 rule Sol:For 8 strips ,the values of y=sinx are tabulated as follows :

x 08 4

38 2

58

34

78

sinx 0 0.3827 0.7071 0.9239 1 0.9239 0.7071 0.3827 0

(i)Trapezoidal rule

0



1

01

dxx 2

11x 2

1

01

dxx2

h3

31

01

dxx2

3h8

13

1

0

1dx(x 2

t

sin xdx = [4(0.38270 16 0.7071 0.9239) 2]

=1.97425

(ii)Simpson’s 1/3 rule

sin xdx = [4(0.38270 24

=2.0003

0.23 0.923 03827) 2(0.7071 1 0.7071)]

2..Find

in each case. Sol:

by using Simpsons 1/3 and 3/8 rule. Hence obtain the approximate value of π

yWe divide the range (0,1) into six equal parts, each of size h=1/6.The values of point of subdivision are as follows.

at each

x 0 1/6 2/6 3/6 4/6 5/6 1y 1 0.9730 0.9 0.8 0.623 0.5902 0.5

By Simpson’s 1/3 rule,

[(y0 y6 ) 4( y1 y3 y5 ) 2( y2

y4 )]

1 [1.5

18 4(2.3632) 2(1.5923)]

=0.7854 (1)By Simpson’s 3/8 rule,

[(y0 y6 ) ( y1 y2 y4 y5 ) 2 y3 ]

1 [1.5

16 (3.1555) .6]=0.7854 (2)

But tan 1 x)

1 an 1 (1) tan 1 (0)

2

2 2



1 dx0 1x

From equation (1) and (2), we have ,π=3.1416

Exercise

1.Evaluate

Ans:1.2

1.2

0

e x dx using (i) Simpson’s 1/3 rule (ii) Simpson’s 3/8 rule ,taking h=0.2

1.2

(i)0e x dx =0.80675 (ii) e x dx =0.80674

0

2. Evaluate cosxdx by dividing the interval into 8 strips using (i) Trapezoidal rule0

(ii) Simpson’s 1/3 rule

Romberg’s method Problems

1. Use Romberg’s method, to compute I correct to 4 decimal places. Hence find

log e 2Sol:The value of I can be found by using Trapezoidal rule with=0.5, 0.25, 0.125

(i) h=0.5 the values of x and y are tabulated as below:

x 0 0.5 1y 1 0.6667 0.5

Trapezoidal rule givesI

1

[1.54

2(0.6667 )]

=0.7084(i) h=0.25, the tabulated values of x and y are as given below:

x 0 0.25 0.5 0.75 1y 1 0.8 0.6667 0.5714 0.5

Trapezoidal rule givesI

1 [1.5

8=0.6970

2(0.8 0.6667 0.5714)]



(ii) h=0.125 , , the tabulated values of x and y are as given below:

e 0



I 2I1

3I

x 0 0.125 0.250 0.375 0.5 0.625 0.750 0.875 1y 1 0.8889 0.8 0.7273 0.6667 0.6154 0.5714 0.5333 0.5

Trapezoidal rule gives,I

1 [1.5

16=0.6941

2(0.8889 0.8 0.7273 0.6667 0.6154 0.5714 0.5333)]

We have approximations for I as I1=0.7084,I2=0.6970,I3=0.6941Use Romberg’s method,the better approximation are caluculated as follws

Using I2

I (h, h

)2

0.6932

I ( h

, h

)0.6931

2 4I (h,

h , h

)0.6931

2 4By actual integration,

1

Idx log (1 x)1 log 2

0 1 x

log e 2 =0.6931

2.Use Romberg’s method ,evaluate

Ans:I=1.9990Exercise:

sin xdx ,correct to four decimal places.0

1.Use Romberg’s method to compute correct to 4 decimal places.Hence find an approximate value ofAns:3.1416

3.3 Gaussian Quadrature

For evaluating the integral Ib

f (x)dx ,we derived some integration rulesa

which require the values of the function at equally spaced points of the interval. Gauss derived formula which uses the same number of function values but with the different spacing gives better accuracy.

Gauss’s formula is ex pressed in the



..

n

wi

i 1

form 1

F (u)du

1

w1 F (u1

)w2 F (u2

).... wn F(un )

= F (ui )



[ f

f

)

Wherewi and u i are called the weights and abscissa respectively. In this formula, the abscissa

and weights are symmetrical with respect to the middle of the interval. The one-point Gaussian Quadrature formula is given by1

f (x)dx1

2 f (0) , which is exact for polynomials of degree upto 1.

Two point Gaussian formula1

f (x)dx1

f (1 ) 3

( 1

)]3

And this is exact for polynomials of degree upto 3.

Three point Gaussian formula1

f (x)dx1

8 f (0)

9

5 [ f (

3

)9 5

( 3

)]5

Which is exact for polynomials of degree upto 5

Note:

Number of terms

Values of t Weighting factor

Valid upto degree

2

1 1T= &

3 3-0.57735&

0.57735

1&1 3

33

0.77465

30.7746

5

50.5555

98

0.888995

0.55559

5

Error terms

The error in two –point Gaussian formula= 1135 f iv ( and the error in three point Gaussian

formula is= Note 1



)

15750f vi (

0



1

11

dxx2

13

13f[

1

11

dxx2

35

35f

1

11

dxx2

1

1 x2 dx

11 x4

x 2 dx1 x 4

35

35f

The integralb

F (t)dt ,can be transformed into

a

1

f (x)dx by h line transformation1

Problems

t1

(b2

a)x1

(b a)2

1.Evaluate by two point and three point Gaussian formula and compare with the

exact value.

Sol:

By two –point Gaussian formula,

1

f (x)dx f (

1

) ( )]

=1.5

By three-point Gaussian formula

1

f (x)dx1

8 f (0)

9

5 [ f

(9

) ( )]

=1.5833

1 dxBut exact value= 2

0 1 x2[tan 1 x]1

2.5708

2.Evaluate by using Gaussian three point formula

Sol:

Here f (x)By three-point Gaussian formula

1

f (x)dx1 8

f (0) 9

2



5 [ f (

9 ))]

…………(1)

35

35

35

35f



1 x2 dx

11 x4

2 dx

1 x

t32

2 dx

1 x

1 2dt 1 dt1 (t3)21 (t3)

13

13f[

1 1313133

f ( )15 34

f ( )15 34

By equation (1),

1

f (x)dx1

8 f (0)

9

= 2521

5 [ f

(9

) ( )]

=0.4902

3.Use Gaussian two point formula ,to evaluate

Sol: a=1,b=2.

The transformation is x1

(b2

x

a)t1

(b a)2

=

By two –point Gaussian formula,

1

f (x)dx f (

1

=

) ( )]



=0.6923

Exercise



31 0

f

f

f

f44hk9

[f

4f

4f

1. Obtain two point and three point Gaussian formula for the gauss-chebyshev Quadrature1 f (x)

formula, given by I dx2

Ans:

1 1 x

; x13 2

1

2. Find I xdx ,by Gaussian three point formula, correct to 4 decimal places.0

Ans: 0.5

3. Evaluate I1

cosxdx1

using Gaussian two point and three point formula

Ans: 1.68311

4. Evaluate I (1 x1

x2 )dx using Gaussian here point formula. Ans: 8/3

3.4 Double integrals

In this chapter we shall discuss the evaluation of

Trapezoidal and Simpson’s rule.

b d

f (x, y)dxdy usinga c

The formulae for the evaluation of a double integral can be obtained by repeatedly applying the Trapezoidal and simpson’s rules.

ym xn

Consider the integral Iy0 x0

f (x, y)dxdy (1)

The integration in (1) can be obtained by successive application of any numerical integration formula with respect to different variables.Trapezoidal Rule for double integral

Ihk4

[ f (x0 , y0 )2[ f (x1 , y0

)f (x2 , y0 )

2 f (x0 , y1 )2 f (x1 , y1

)2 f (x2 , y1 )

(x0 , y2 )](x1 , y2 )]

(x2 , y2 )

Simpson’s rule for double integralThe general expression for Simpson’s rule for the double integral (1) can

also be derived.In particular, Simpson’s rule for the evaluation of

f (xi 1 , y j 1 )

I(xi 1 , y j 1 )

f (xi 1 , y j )

f (xi 1 , y j )

(x0i 1 , y j 1

)](xi 1 , y j 1 )

[ f (xi , yij 1 ) f (x j , y j

)(xi , y j 1 )]



Problems:



ff

f

hk [9f f

f

1 1

0 0 1dxdy

x y

1xy

1. Evaluate I

Sol:

1 1

ex y dxdy0 0

using trapezoidal and Simpson’s rule

Taking h=k=0.5The table values of ex y are given a follows

(x0) (x1) (x2)y\x 0 0.5 1

0 1 1.6487 2.7183

0.5 1.6487 2.7183 4.4817

1 2.7183 4.4817 7.3891

(i) Using Trapezoidal rule,we obtainy2 x2

Iy0 x

Ihk4

f (x, y)dxdy

[ f (x0 , y0 )2[ f (x1 , y0

)f (x2 , y0 )

2 f (x0 , y1 )2 f (x1 , y1

)2 f (x2 , y1 )

where

(x0 , y2 )](x1 , y2 )]

(x2 , y2 )

f (x, y) = ex y

=3.0762(ii) Using Simpson’s rule,we obtain

I

=2.9543

f (x0 , y0

)(x2 , y0 )

4 f (x0 , y1

)4 f (x2 , y1 )

(x0 , y2 )](x2 , y2 )]

4 f (x1 , y0

)16 f (x1 , y1

)4(x1 , y2 )]

2.Evaluate I by (i) Trapezoidal rule and (ii) Simpson’s rule with step sizes h=k=0.5

Sol:Taking h=k=0.5

1The table values of are given a follows

(x0) (x1) (x2)y\x 0 0.5 1

0 1 0.67 0.5

0.5 0.67 0.5 0.4

1 0.5 0.4 0.33

y2 x2

y0 x



1xy

f

f

f

hk [9f f

f

2 2 dxdy

1 1 xy

(i) Using Trapezoidal rule,we obtain1

I f (x, y)dxdy where f (x, y) =

Ihk4

[ f (x0 , y0 )2[ f (x1 , y0

)f (x2 , y0 )

2 f (x0 , y1 )2 f (x1 , y1

)2 f (x2 , y1 )

(x0 , y2 )](x1 , y2 )]

(x2 , y2 )

=0.5381

(i) Using Simpson’s rule,we obtain

I

=0.5247

f (x0 , y0

)(x2 , y0 )

4 f (x0 , y1

)4 f (x2 , y1 )

(x0 , y2 )](x2 , y2 )]

4 f (x1 , y0

)16 f (x1 , y1

)4(x1 , y2 )]

Exercise

1. Compute I1 1

xydxdy with h=0.25 and k=0.5 using Simpson’s rule0 0

Ans:I=0.25

2. Using Trapezoidal rule ,evaluate I taking four sub intervlas

Ans:I=0.34065



Tutorial problems

Tutorial -1

1. Find the value of f (3) using divided differences, given data

x 0 1 4 5F(x) 8 11 68 123

Ans:162. Find the first and second derivatives of the function at the point x=1.2 from the following data

X 1 2 3 4 5Y 0 1 5 6 8Ans:14.17

3.A rod is rotating in a plane .The angle θ (in radians) through which the rod has turned for various values of time t (seconds) are given below

t 0 0.2 0.4 0.6 0.8 1.0 1.2θ 0 0.122 0.493 1.123 2.022 3.220 4.666

Find the angular velocity and angular acceleration of the rod when t=06 seconds.

Ans:6.7275 radians/sec2.

4.From the table given below ,for what value of x, y is minimum? Also find this value of y.

X 3 4 5 6 7 8Y 0.205 0.240 0.25 0.262 0.250 0.224

Ans:5.6875,0.2628

5.Find the first and second derivatives of y at x=500 from the following dat

x 500 510 520 530 540 550y 6.2146 6.2344 6.2538 6.2792 6.2196 6.3099Ans:0.002000,-0.0000040

1.Evaluate1 xdx

Tutorial -2

by Simpson’s 1/3 rule with h=0.1

2.Evalute

0 1 x1

cos2 xdx with h=0.1 by Simpson’s 1/3 rule.0

1

3.Apply Trapezoidal and Simpson’s rules ,to find 1 x 2 dx ,taking h=0.1.0

6 dx0 1x



7 dx

1 x

1

11

dxx 4

1 cos2xdx1 1sin x

4.Evaluate by dividing the range into eight equal parts.

5.Using Simpson’s rule ,find log e 7approximately from the integral I

1 dxTutorial -3

1. Evaluate using Gaussian three point formula.2

2. Eva

luate

3.Evalu

ate

1 1 x1

ex dx using three-term Gaussian Two-point Quadrature formula.0

using Gaussian three point quadrature formula.

4. Evaluate using Gaussian two point quadrature formula.

5.Evaluate4

(x 2

2

2x)dx using Gaussian two point quadrature formula.

Question Bank

Part A

1. State the disadvantages of Taylor series method.Sol:

In the differential equation = f(x,y) the function f(x,y) may have a complicated

algebraical structure. Then the evaluation of higher order derivatives may become tedious. This is the demerit of this method.

2. Which is better Taylor’s method or R.K method?Sol:

R.K methods do not require prior calculation of higher derivatives of y(x), as the Taylor method does. Since the differential equations using in application are often complicated, the

calculation of derivatives may be difficult. Also R.K formulas involve the computations of f(x,y) at various positions instead of derivatives and this function occurs in the given equation.



3. What is a predictor- collector method of solving a differential equation?Sol:



predictor- collector methods are methods which require the values of y at xn, xn-1, xn-2,… for computing the values of y at xn+1. We first use a formula to find the values of y at xn+1 and this is known as a predictor formula. The value of y so get is improved or corrected by another formula known as corrector formula.

4. Define a difference Quotient.Sol:

A difference quotient is the quotient obtained by dividing the difference between two values of a function, by the difference between the two corresponding values of the independent.

dy d 2 y5. Write down the expression for &

dx dx2 atx xn by Newton’s backward difference formula.

6. State the formula for Simpson’s 3/8th rule.7. Write Newton’s forward derivative formula.

8.State the Romberg’s method integration formula fin the value of Ib

f (x)dx using h and h/2.a

9.Write down the Simpson’s 3/8 rule of integration given (n+1) data.4

10.Evaluate f (x)dx from the table by Simpson’s 3/8 rule.1

Part B

1.Apply three point Gaussian quadrature formula to evaluate .

2.Using Trapezoidal rule, evaluate numerically with h=0.2 along x-direction and

k=0.25 along y-direction.

3.Find the first and second derivative of the function tabulated below at x=0.6

x 0.4 0.5 0.6 0.7 0.8y 1.5836 1.7974 2.0442 2.3275 2.6511

4. Using Romberg’s method to compute dx correct to 4 decimal places.Also evaluate

the same integral using three –point Gauss quadrature formula. Comment on theobtained values by comparing with exact value of the integral which is equal to .

5.Evaluate using Simpson’s rule by taking h= and k =

. 6.Find and at x=51 from the following data.

x 50 60 70 80 90y 19.96 36.65 58.81 77.21 94.61



7.Evaluate I= by dividing the range into ten equal parts,using (i)Trapezoidal rule(ii)Simpson’s one-third rule. Verify your answer with actual integration.

8. Find the first two derivatives of at x=50 and x=56, for the given table :x 50 51 52 53 54 55 56

Y=x1/3 3.6840 3.7084 3.7325 3.7325 3.7798 3.8030 3.8259

9.The velocities of a car running on a straight road at intervals of 2 minutes are given below:

Using Simpson’s 1/3 rd- rule find the distance covered by the car.

10. Given the following data, find y’(6) and the maximum value of y (if it exists)

x 0 2 3 4 7 9Y 4 26 58 112 466 922

Time(min) 0 2 4 6 8 10 12Velocity(km/hr) 0 22 30 27 18 7 0



Chapter 4

Numerical Solution of ordinary Differential equations

Taylor’s series method

Introduction

An ordinary differential equation of order n is a relation of the formr

F (x, y, y , y ,...... y(n)

) 0 where y=y(x) and y (r

)

d y .The solution of this differential equation

dxr

involves n constants and these constants are determined with help of n conditions.

Single step methods

In these methods we use information about the curve at one point and we do not iterate the solution. The method involves more evaluation of the function. We will discuss the Numerical solution by Taylor series method, Euler methods and Runge - Kutta methods all require the information at a single point x=x0.

Multi step methods

These methods required fewer evaluations of the functions to estimate the solution at a point and iteration are performed till sufficient accuracy is achieved. Estimation of error is possible and the methods are called Predictor-corrector methods. In this type we mainly discuss Milne’s and Adams-Bashforth method.

In the multi step method , to compute yn+1,we need the functional values yn,yn-1, yn-2 and yn-3.

4.1 Taylor Series method

Consider the first order differential equation dy

dx f (x, y), y(x0

)y0 ……(1)

The solution of the above initial value problem is obtained in two types

Power series solution Point wise solution

(i) Power series solution



+……. (2)

0



d r ydxr

d r ydxr

Where y ( r ) at(

Using equation (1) the derivatives can be found by means of successive differentiations .Expressions (2) gives the value for every value of x for which (2) converges.

(ii)Point wise solution

= + h

+ + +…1!

Where (r )m at (

Where m=0,1,2,…….

Problems:

1. Using Taylor series method find y at x=0.1

if Solution:

Given = y-1 and =0, =1,h=.1

Taylor series formula for is

= + h

+ + +…1!

dy = y-1,y(0)=1.

dx

= y-1 = =0-1=-1

=2xy+ + =0+0=0

= + +2x

=2x +2y+ +2x

=2y+ 4x +

=2 +4 +

=2(1)+4(0)(-1)+( )(0)

=2

y



= +4 + +

= +4x +4 + + 2x

= + 6x +

= +6 +

=6(-1)+6(0)(0)+(0)(2)

=-6



(1) =1+ (-1)+ (0)+ (2)+ +…

Y(0.1)=1-(0.1)+ +…

=1-0.1+0.00033-0.000025=0.900305

2. Find the Taylor series solution with three terms for the initial value problem. = +y,y(1)=1.

Solution:

Given = +y, =1, =1.

= +y = + =1+1=2

=3 + +

=3(1)+2

=5

=6x+ = +

=6(1)+5

=11

Taylor series formula

Y= +…….

=1+(x-1)(2)+ (5)+ (11)(app)

=1+2(x-1)+ +

Exercise:

1.Find y(0.2),y(0.4) given =x +1,y(0)=1

[Ans:-1.226,1.5421 ]

y (


100

CIVIL ENGINEERING

hfh

y

y

2.Find y(0.1) given = y-1,y(0)=1

[Ans:- 0.9003 ]

EULER’S METHOD:

Introduction

In Taylor’s series method, we obtain approximate solutions of the initial value problem

dyf (x, y), y(x0 )

dxy0 as a power series in x , and the solution can be used to compare y

numerically specified value x near x0.

In Euler’s methods,we compute the values of y for xi x0 ih,i 1,2..... with a step h>0

i.e., yi y(xi ) where x

x0 ih,i 1,2.....

Euler’s method

= +f ),n=0,1,2,3,……

Modified Euler’s methodTo compute for

(i) n 1 yn hf (xn , yn )

(ii) yn 1 yn [ f (xn2, yn ) (xn , yn 1 (1) ) error n=0,1,2…..

Error=O(h3)

Note

In Euler method yn 1 yn

Where y hf (x0 , y0 )

Where f (x0 , y0

)slope

at

f (x0 , y0 ) .

In modified Euler’s method


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y

yn 1 yn ,

Where =h[average of the slopes at x0 and x1](Or)


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1

=h[average of the values of dy at the ends of the interval x to x ]

dx 0 1

1.Using Euler’s method compute y in the range 0 .5 if y satisfies 3x+ ,y(0)=1.

Solution:

Here f(x,y) = 3x+ , =0, =1By Euler’s method

= +f ),n=0,1,2,3,…… (1)Choosing h=0.1, we compute the value of y using (1)

= +f

= +hf(

),n=0,1,2,3,……

)=1+(0.1)[3(0)+ ] =1.1

= +hf( )=1.1+(0.1)[0.3+ ] =1.251

= +hf( )=1.251+(0.1)[0.6+ ] =1.4675

=

=

+hf(

+hf(

)=1.4675+(0.1)[0.9+

)=1.7728+(0.1)[1.2+

] =1.7728

] =2.2071

Exercise:

1.Given that dy

dx x y and y=1 when x=0,Find y when x=0.05,0.10 and 0.15 using modified

Euler method.

Ans:

1.0525,1.1103,1.1736

2.Using Euler’s method,solve y x y xy, y(0)

.Compute y(1.0) with h= 0.2

Ans:3.9778

MODIFIED EULER’S METHOD:

1. Given =-x , y(0)=2. Using Euler’s modified method, find y(0.2)in two steps of 0.1 each.

Solution:


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Let = +h. here , and h=0.1


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The value of y at x= is

By modified Euler method,

= hf(

f( f( ]

Here f(x,y) = -x

f( = - =0

2+(0.1)(0)=2

2+ [f(0,2)+f(0.1,2)]

=2+0.05[0+(-0.1)(4)]

1.98

Y(0.1)=1.98

To find y(0.2)

=1.98 –(0.1)(0.1)(1.98)=1.9602

+ f( f( ]

=1.9602-(0.05)[(0.198)+(0.39204)]

Hence y(0.2)=1.9307

Runge-kutta method

The Taylor’s series method of solving differential equations numerically is restricted because of the evaluation of the higher order derivatives.Runge-kutta methods of solving intial value problems do not require the calculations of higher order derivatives and give greater


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h k1

h k2

kh

22 k

1x

accuracy.The Runge-Kutta formula posses the advantage of requiring only the function values at some selected points.These methods agree with Taylor series solutions upto the term in hr where r is called the order of that method.

Fourth-order Runge-Kutta method

This only is commonly used for solving initial values problemdy

f (x, y), y(x0 ) y0dx

Working rule

The value of y1 y(x1 ) where x1

x0 h where h is the step-size is obtained as follows.We

calculate successively.

k1 hf (xm , ym )

k2 hf (xm 2 , ym 2

)

k hf (x , y )3 m 2 m 2

k4 hf (xm , ym 3 )

Finally compute the increment

y y1

[k k k] where m=0,1,2……

m 1 m 6 1 2 3 4

Problems

1. Obtain the value of y at x =0.2 if satisfies dy

dxmethod of fourth order.

Sol:

x2 y , x0

0, y0

using Runge-kutta

Heref (x, y) x2 y x,

x0

0, y0 1

x1Let

x0 h

Choosing h=0.1,x1=0.1


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Then by R-K fourth order method,

k22


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h k1

h

h

k2

k

k22

h

h

kh

k1

k2

1

y y1

[k k k ]1 0 6 1 2 3 4

k1 hf (x0 , y0 ) 0

k hf (x , y ) 0.005252 0 2 0

k3 hf (x0 2 ,

y0

2

) 0.005252

k4 hf (x0 , y0 3 ) 0.0110050

y(0.1)=1.0053

To find y2 y(x2 ) where x2

x1 h Taking x2=0.2

y y1

[k k k ]2 1 6 1 2 3 4

k1 hf (x1 , y1

)0.0110

k2 hf (x1

k3 hf (x1

, y1

2

2 ,

y1

) 0.017272

) 0.017282

k4 hf (x1 , y1

3 ) 0.02409

y(0.2)=1.0227

2.Apply Runge –kutta method to find an approximate value of y for x=0.2 in steps of 0.1 if

dyx dx y 2 ,

y(0),correct to four decimal places.

Sol:

Here f(x,y)=x+y2,x0=0,y(0)=1

Then by R-K fourth order method,


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k22y y1

[k k k ]1 0 6 1 2 3 4


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h k1

h

h

k2

0k

k22

h

h

kh

k1

k2

1

1

k1 hf (x0 , y0 ) 0.1

k hf (x , y ) 0.11522 0 2 0

k3 hf (x0 2 ,

y0

2

) 0.11682

k4 hf (x0 , y0 3 ) .1347

y(0.1)=1.1165

To find y2 y(x2 )

where x2 x1 h Taking x2=0.2

y y1

[k k k ]2 1 6 1 2 3 4

k1 hf (x1 , y1

)0.1347

k2 hf (x1

k3 hf (x1

, y1

2

2 ,

y1

) 0.11512

) 0.15762

k4 hf (x1 , y1

3 ) 0.1823

y(0.2)=1.2736

Exercise:

1.Use Runge-kutta method to find when x=1.2 in steps of 0.1 ,given that dy

dx x2 y 2 , y(1) .5

Ans:1.8955, 2.5043

2.Solve y xy for x=0.2(0.2)0.6 by Fourht order method,given that y=2 when x=0

Ans: 2.2431,2.5886, 3.0719


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h5

Multi step methods(Predictor-Corrector Methods)

Introduction

Predictor-Corrector Methods are methods which require function values atxn , xn 1 , xn 2 , xn 3 for the compulation of the function value at xn+1.A predictor is used to find the value of y at xn+1 and then a corrector formula to improve the value of yn+1.

The following two methods are discussed in this chapter.

Milne’s method Adam’s method.

Milne’s Predictor-Corrector method

Milne’s predictor formula is

14h5

And the error =45 y (v) ( )

Where xn 3 xn 1

Milne’s corrector formula is

And the error =y (v) )90

Where xn 1 xn 1


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Problems1.Using Milne’s method ,compute y(0.8) given thatdy

1dxy2 , y(0)

1, y(0.2)

0.2027, y(0.4)

0.4228, y(0.6)

0.6841

Sol:We have the following table of values

X y Y’=1+y2

00.20.40.6

00.20270.42280.6841

1.01.04111.17871.4681

x0 0, x1

0.2, x2

0.4, x3

0.6

y0 0, y1

y0 1, y1

0.2027, y2

1.0411, y2

0.4228, y3

1.1787, y3

0.6841

1.4681

To find y(0.8)x8 0.8 ,here h=0.2

Milne’s predictor formula iswhere ).

=1.0239=2.0480

Milne’s corrector formula iswhere ).

=1.0294 y(0.8)=1.0294


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2. Compute y(0.4) by Milne’s method ,given that ymethod to find the starting values.

x y, y(0)

with h=0.1 use Taylor’s

) )


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Ans:1.9816, 1.5774

Adams-bashforth Predictor and Adams-bashforth corrector formula.

Adams-bash forth predictor formula is

Adams –bash forth corrector formula iswhere ).

The error in these formulas are 251h5 f iv ( 720 and 19

720 h5 f iv (

Problems

1.Given y y x2 , y(0)

1, y(0.2) 1.2186 , y(0.4) 1.4682 , y(0.6) 1.7379 estimate y(0.8) by

Adam’s-Bashforth method.

Sol:

x0 0, x1

0.2, x2 0.4, x3 0.6

y0 0, y1

y0 1, y1

1.2186, y2

1.1782, y2

1.4682, y3

1.3082, y3

1.7379

1.3779

To find y4 for x4 0.8 here h=0.2

By Predictor formula,

=2.0146


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1y4 .3746


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=2.0145

y(0.8)=2.0145

2.Given y

1 y 2 , y(0)

0, y(0.2) 0.2027 , y(0.4) 1.4228 , y(0.6)

0.6841 ,Estimate y(0.8)

using Adam’s method.

Sol:

x0 0, x1

0.2, x2

0.4, x3

0.6

y0 0, y1

y0 1, y1

0.2027, y2

1.0411, y2

0.4228, y3

1.1786, y3

0.6841

1.4680

To find y4 for x4 0.8 here h=0.2

By Predictor formula,

=1.0235

y4 2.0475

=1.0297

y(0.8)=1.0297


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y

1xyx 2

1

1

1

Tutorial problems

Tutorial-1

dy1.Using Taylor\s series method ,find y(0.2) approximately ,given that

dxcorrect to four decimal places.

2 y 3ex y(0)=0

dy2.Compute y for x=0.1,0.2 correct to four decimal places given

dxseries method.

y x, y(0) 2 using taylor’s

3.Using Euler’s Modified method ,find the values of y at x=0.1,0.2,0.3,0.4 given

dy1dx y, y(0) 0 correct to four decimal places.

dy4. Using Euler’s method ,find y(0.1),y(0.2) and y(0.3)given that 1

dx

dy5.Using Euler’s modified method, find a solution of the equation

dx

y 2 , y(0)

x

0

with the initial

condition y(0)=1 for the range 0 x 0.6 in steps of 0.2.

Tutorial-2

1.Use R-K fourth order method, of find y at x=0.1(0.1)0.3 if y , (0)

dy2.Apply R-K method of order 4 to solve

dx xy1 3 , y(1) with h=0.1

3.Using R-K method to solve

d 2 y dy x y 2 , y(0)

1, y (0) 0 to find y(0.2) and

y (0.2)

dx2 dx

4.Use the R-K method of fourth order ,find y(0.1)=1 given that2

d yx 2 dy 2xy 1, y(0) 1, y (0) 0

dx2 dx

dy5.Compute y and z for x=0.1 given

dx1 z; dz

dxx y, y(0)

0, z(0)


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111

1

Tutorial-3

1.Apply Milne’s method ,to find y(0.4) given that ymethod to compute y(0.1),y(0.2),y(0.3).

xy y 2 , y(0)

1. Use Taylor’s series

dy2. using Adam-Bashforth method, find the solution of

dxy(0)=1,,y(0.1)=1.1103,y(0.2)=1.2428,y(0.3)=1.3997

3. Using adam’s method ,determine y(0.4) given that

x y at x=0.4,given the values

dy 1 xy,

y(0) dx 2 1, y(0.1)

.0025, y(0.2)

.0101, y(0.3)

.0228

4. Given y1

x 2 y,

y(0)2

, computey(0.1), y(0.2), y(0.3) by the fourth order R-K method and

y(0.4) by Adam’s method.

5. Compute y for x=0.1(0.1)0.3 by the fourth order R-K method and y(0.4) by Adam’s method ,ifdy

1 x dx

4 y, y(0) 1


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Question Bank

Part A

1.State Modified Euler algorithm to solve y’= ,y(x0)=y) at x=x0 +h

Sol:

2. State the disadvantage of Taylor series method.Sol:

In the differential equation f (x, y), = f (x, y), the function f (x, y),may have acomplicated algebraical structure. Then the evaluation of higher order derivatives may become tedious. This is the demerit of this method.

3. Write the merits and demerits of the Taylor method of solution.Sol:

The method gives a straight forward adaptation of classic to develop the solution as an infinite series. It is a powerful single step method if we are able to find the successive derivatives easily. If f (x.y) involves some complicated algebraic structures then the calculation of higher derivatives becomes tedious and the method fails.This is the major drawback of this method. However the method will be very useful for finding the starting values for powerful methods like Runge - Kutta method, Milne’s method etc.

4.Which is better Taylor’s method or R. K. Method?(or) State the special advantage of Runge-Kutta method over taylor series method

Sol:

R.K Methods do not require prior calculation of higher derivatives of y(x) ,as the Taylor method does. Since the differential equations using in applications are often complicated, the calculation of derivatives may be difficult.

Also the R.K formulas involve the computation of f (x, y) at various positions, instead of derivatives and this function occurs in the given equation.

5.Compare Runge-Kutta methods and predictor – corrector methods for solution of initial value problem.

Sol:Runge-Kutta methods

1.Runge-methods are self starting,since they do not use information from previously calculated points.


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1

2.As mesne are self starting,an easy change in the step size can be made at any stage. 3.Since these methods require several evaluations of the function f (x, y), they are time consuming.4.In these methods,it is not possible to get any information about truncation error. Predictor Corrector methods:

1.These methods require information about prior points and so they are not self starting. 2.In these methods it is not possible to get easily a good estimate of the truncation error.

6. What is a Predictor-collector method of solving a differential equation?Sol:

Predictor-collector methods are methods which require the values of y at xn,xn-1,xn-2,… for computing the value of y at . x n+1 We first use a formula to find thevalue of y at . x n+1 and this is known as a predictor formula.The value of y so got is improved or corrected by another formula known as corrector formula.

7. State the third order R.K method algorithm to find the numerical solution of the first order differential equation.Sol:

To solve the differential equation y′ = f (x, y) by the third order R.K method, we use the following algorithm.

and8.Write Milne’s predictor formula and Milne’s corrector formula.Sol:

Milne’s predictor formula is

Milne’s corrector formula iswhere ).

where ).

9.Write down Adams-bashforth Predictor and Adams-bashforth corrector formula.Sol:Adams-bashforth predictor formula is

Adams –bashforth corrector formula is

10.By Taylor’s series method ,find y(1.1) given yAns:1.2053

x y, y(1)

where ).


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Part B

1.Using Runge-Kutta method find an approximate value of y for x=0.20 if =x + given that y=1 when x=0.

2.Given that y” +xy’+y =0,y(0)=1,y’(0)=0 obtain y for x=0,1,0.2,0.3 by Taylor’s series method and find the solution for y(0.4) by Milne’s method.

3.Obtain y by Taylor series method,given that y’=xy+1,y(0)=1for x=0.1 and 0.2 correct to four decimal places.4.Solve for y(0.1) and z(0.1) from the simultaneous differential equations

=2y +z: =y-3z; y(0)=0,z(0)=0.5 using Runge-Kutta method of the fourth order.

5.Using Adams method find y(1.4) given y’=x2(1+y) , y(1)=1, y(1.1)=1.233, y(1.2)=1.548 and y(1.3)=1.979.

6.Using Milne’s Predictor-Corrector formula to find y(0.4) given = , y(0)=1,y(0.1)=1.06, y(0.2)=1.12 and y(0.3)=1.21.

7.Using Modified Euler’s method , find y(4.1) and y(4.2) if 5x + -2 =0: y(4)=1.

8.Given that =1+ ; y(0.6)=0.6841, y(0.4)=0.4228, y(0.2)=0.2027, y(0)=0, find y(-0.2) usingMilne’s method.

9.Given that y’=y-x2 ; y(0)=1: y(0.2)=1.1218; y(0.4)=1.4682 and y(0.6)= 1.7379 , evaluate y(0.8) by Adam’s predictor –Corrector method.10.Solve by Euler’s method the followingdifferential equation x=0.1, correct to four decimal places, = with initial condition

y(0)=1.


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2 yi 1 yi 2 ]1

Chapter 5

Boundary value Problems in ODE&PDE

5.1 Solution of Boundary value problems in ODE

Introduction

F(x, y, y , y )The solution of a differential equation of second order of the form

0 contains two arbitrary constants. These constants are determined by means oftwo conditions. The conditions on y and or their combination are prescribed at two different values of x are called boundary conditions.

The differential equation together with the boundary conditions is called aboundary value problem.

In this chapter ,we consider the finite difference method of solving linear boundary value problems of the form.

Finite difference approximations to derivatives

First derivative approximations

y (x)

y (x)

y(x

y(x)

h) h

y(x h

y(x)

h)

O(h)

O(h)

(Forward difference)

(Backward difference)

y (x) y(x h) y(xh) 2h

O(h2 ) (Central difference)

Second derivative approximations

y (x) y(x h) 2 y(x)

h2

y(x h)O(h2 ) (Central difference)

Third derivative approximations

yiii (x)2h3 [ yi 2 2 yi 1


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Fourth derivative approximations


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11 yi 2 ]6 yi4 y

yi 1yi 1

2h

yi 1 2 yi yi 1

1

yiv (x)4 [ yi 2

h4 yi 1 i

Solution of ordinary differential equations of Second order

yi

yi h2

Problems

1. Solve xy y 0, y(1)

, y(2) 2 with h=0. And h=0.25 by using finite difference method.

Sol:

(i) Divide the interval[1,2] into two sub-intervals with h=(2-1)/2=0.5

X0=1 x1=1.5 x2=2

Let xi 1 xi h ,

Here x0 1, x1

1.5, x2 2

By Boundary conditions arey(1) 1, y(2) 2

y0 1, y2 2

We have to determiney1 for x1 =1.5


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yi 1 2 yi

h2

yi 1

xi

h2[ yi 1 2 yi

Replacing the derivative y by y

yi 1 ] yi 0 (1)


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18

Put i=1 we get 4x1[ y0

2 y1

y2 ] y1 0

Using x1 =1.5, y0 1, y2 2 we get, y111

y(1.5)=1.6364

(ii) With h=0.25 ,we have the number of intervals =4

X0=1 x1=1.25 x2=1.5 x3=1.75 x4=2

By boundary condition, y0 1, y4 2

We find the unknown from the relation,

16 xi ( yi 12 yi

yi 1 ) yi 0, i

1,2,3.... (2)

Put i=1 in (2), 39 y1 20 y2 20

Put i=2 in (2), 28 y147 y2 24 y3 0

Put i=3 in (2), 28 y2 55 y3 56 0

We have the system of equations ,

39 y1 20 y2 20

28 y147 y2 24 y3 0

28 y2 55 y3 56


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Solution of Gauss-Elimination method


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1

yk

=

The equivalent system isy1 0.513y2 0.513

1.445y2 y3 0.5131.838y3 3.403

By back substitution, the solution is y3 1.851 , y2

1.635 , y1

1.351

Hence y(1.25)=1.351,y(1.5)=1.635,y(1.75)=1.851

Exercise

1. Solve by finite difference method, the boundary value

problem and y (2) =4.

d 2 ydx2

y 0 with y (0) =0

Ans: y3 2.3583 , y2 1.3061 , y1 0.58052. Solve xy y 0, y

(1)0andy(2) with h=0.5

5.2 Solution of Laplace Equation and Poisson equation

Partial differential equations with boundary conditions can be solved in a region by replacing the partial derivative by their finite difference approximations. The finite difference approximations to partial derivatives at a point (xi,yi) are given below. The xy-plane is


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divided into a network of rectangle of lengh x h and breadth

by drawing the lines x=ih

and y=jk, parallel to x and y axes.The points of intersection of these lines are called grid points or


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1 uui 1, u

1ui 11ui 1,1ui 1,

bbb

bbu

mesh points or lattics points.The grid points (xi , y j ) is denoted by (i,j) and is surrounded by the

neighbouring grid points (i-1,j),(i+1,j),(i,j-1),(i,j+1) etc.,

Note

The most general linear P.D.E of second order can be written as

2u 2uA B2u uC D E

uFu f (x, y) (1)

x2 x y y 2 x y

Where A,B,C,D,E,F are in general functions of x and y.

The equation (1) is said to be

Elliptic if B2-4AC<0 Parabolic if B2-4AC=0 Hyperbolic if B2-4AC>0

Solution of Laplace equation uxx+uyy=0

ui, j1

[u4

i 1, j j i, j

i, j 1 ]

This formula is called Standard five point formula

ui, j1

[u4

i 1, j j j , j 1 ]

This expression is called diagonal five point formula.

Leibmann’s Iteration Process

We compute the initial values of u1 ,u2 ,.....u9 by using standard five point formula and

diagonal five point formula .First we compute u5 by standard five point formula (SFPF).

u1

[b ]5 4 7 15 11 3

We compute u1 ,u3 ,u7 .u9 by using diagonal five point formula (DFPF)

u1

[b ]


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bbb

1 4 1 5 3 15

u1

[u3 4 5 5 3 7 ]

bbu

ubb


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uub

ubu

u

b

bu

uu

u1

[b ]7 4 13 5 15 11

u1

[b ]9 4 7 11 9 5

Finally we compute u2 ,u4 ,u6 ,u8 by using standard five point formula.

u1

[u ]2 4 5 3 1 3

u1

[u ]4 4 1 5 15 7

u1

[u6 4 3 9

u1

[u

5 7 ]

]8 4 7 11 9 5

The use of Gauss-seidel iteration method to solve the system of equations obtained by finite difference method is called Leibmann’s method.

Problems

1. Solve the equation

2u 0 for the following mesh,with boundary vaues as shown using

Leibmann’s iteration process.

u1 u2 u3

u4 u5 u6

u7 u8 u9

0 500 1000 500 0

Sol:

Let u1 , u2 …… u9 be the values of u at the interior mesh points of the given region.By symmetry about the lines AB and the line CD,we observeu1 u3 u1 = u7


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u2 u8 u4 u6


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1

u3 u9 u7 u9

u1 = u3

u7 = u9 , u2

u8 , u4 u6

Hence it is enough to find u1 , u2 , u4 , u5

Calculation of rough valuesu5 =1500

u1 =1125

u2 =1187.5

u4 =1437.5Gauss-seidel scheme

u1

[1500 u1 4 2

u4 ]

u1

[2u u 000]2 4 1 5

u1

[20004 4

u1

[2u

u5

2u ]

u4 ]

5 4 2 4

The iteration values are tabulated as followsIteration No k

u1 u2 u4 u5


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0123456789101112

15001031.251000968.75953.1945.3941.4939.4938.4937.9937.7937.6937.6

112511251062.51031.251015.31007.81003.91001.91000.91000.41000.21000.11000.1

1187.513751312.51281.251265.61257.81253.91251.91250.91250.41250.21250.11250.1

1437.512501187.51156.251140.61132.81128.91126.91125.91125.41125.21125.11125.1

u1 = u3

u7 = u9 =937.6, u2

u8 =1000.1, u4

u6 =1250.1, u5 1125 .1


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2u2ux2y 2

ui, j 1 4uui 1

0 0

0 J=3

0 J=2

J=1

1

2.When steady state condition prevail,the temperature distribution of the plate is represented by Laplace equation uxx+uyy=0.The temperature along the edges of the square plate of side 4 are given by along x=y=0,u=x3 along y=4 and u=16y along x=4,divide the square plate into 16 squaremeshes of side h=1 ,compute the temperature at all of the 9 interior grid points by Leibmann’s iteration process.

Solution of Poisson equation

An equation of the type 2u f (x, y) i.e., is called

f (x, y) poisson’s equation

where f(x,y) is a function of x and y.

ui 1, j , j i, j h2 f (ih, jh)

This expression is called the replacement formula.applying this equation at each internal mesh point ,we get a system of linear equations in ui,where ui are the values of u at the internal mesh points.Solving the equations,the values ui are known.

Problems

1.Solve the poisson equation 2u 10(x2 y 2

10) over the square mesh with sides

x=0,y=0,x=3,y=3 and u=0 on the boundary .assume mesh length h=1 unit.

Sol:

A u1 u2 B

0 u3 u4

0 C D

I=0 i=1 i=2 i=3 x

Here the mesh length x h


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CIVIL ENGINEERING

ui 1

Replacement formula at the mesh point (i,j)

ui 1, j, j ui, j 1 ui, j 1 4ui,

j

10(i2 j2 10) (1)


137

CIVIL ENGINEERING

u

ut

2 2ux 2

1 ui 1,

ui, j ui 1, j)

uuui,

u2 3

u1 u4

u2 u3

u1 u4

4u1

4u3

4u4

75, u2

150120

15082.5, u3

67.5

2u 2u2.Solve the poisson equationx2 y 2

81xy,0 x 1;0y

1and

u(0,y)=u(x,0)=0,u(x,1)=u(1,y)=100 with the square meshes ,each of length h=1/3.

5.3 Solution of One dimensional heat equationIn this chapter, we will discuss the finite difference solution of one dimensional heat flow

equation by Explicit and implicit method

Explicit Method(Bender-Schmidt method

Consider the one dimensional heat equation

parabolic equation..This equation is an example of

ui, j 1

Where

ui 1, j

k ah2

1 2 ui, j ui 1, j (1)

Expression (1) is called the explicit formula and it valid for 01

2If λ=1/2 then (1) is reduced into

ui, j 1 [ui 1, j j ]2 (2)

This formula is called Bender-Schmidt formula.

Implicit method (Crank-Nicholson method)

ui 1, j 1

2(1 )ui, j 1 ui 1, j 1 ui 1, j 2(1

This expression is called Crank-Nicholson’s implicit scheme. We note that Crank Nicholson’s scheme converges for all values of λ

When λ=1, i.e., k=ah2 the simplest form of the formula is given by1

[uj 1 4

i 1, j 1 i 1, j 1 i 1, j ui 1, j ]

The use of the above simplest scheme is given below.


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CIVIL ENGINEERING

1

ui 1,

ah2

33

0 j

The value of u at A=Average of the values of u at B, C, D, E

Note

In this scheme, the values of u at a time step are obtained by solving a system of linear equations in the unknowns ui.

Solved Examples

1.Solve uxx

assuming2ut

x h

when u(0,t)=0,u(4,t)=0 and with initial condition u(x,0)=x(4-x) upto t=sec

Sol:

By Bender-Schmidt recurrence relation ,

ui, j 11

[u2

i 1, j j ](1)

kFor applying eqn(1) ,we choose 2

Here a=2,h=1.Then k=1

By initial conditions, u(x,0)=x(4-x) ,we have

ui,0i(4 i)i 1,2,3

, u1,0 , u2,0


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CIVIL ENGINEERING

4, u3,0

By boundary conditions, u(0,t)=0,u0=0,u(4,0)=0u4, j

u


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CIVIL ENGINEERING

1

u

u

3u

ut

2 ux 2

Values of u at t=1

ui,1[ui 1,0

2 i 1,0 ]

u1,11

[u2

0,0 2,0 ] 2

u2,1

u3,1

1

[u21

[u2

1,0

2,0

3,0 ]

4,0 ] 2

The values of u up to t=5 are tabulated below.

j\i 0 1 2 3 4

0

1

2

3

4

5

0

0

0

0

0

0

3

2

1.5

1

0.75

0

4

3

2

1.5

1

0.75

3

2

1.5

1

0.75

0.5

0

0

0

0

0

0

2.Solve the equationsubject to the conditions u(0,t)=u(5,t)=0 and

u(x,0) x 2

(25x 2

) taking h=1 and k=1/2,tabulate the values of u upto t=4 sec.


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CIVIL ENGINEERING

Sol:

Here a=1,h=1

For λ=1/2,we must choose k=ah2/2

0 j


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0 u5,

0u0

ii

0 j

K=1/2

By boundary conditions

u(0, t)

u(5, t)

u(x,0)

ui,0

,

j

j

x 2 (252 (25

x 2 )2 ),i 0,1,2,3,4,5

u1,0 24, u2,0

84, u3,0

144, u4,0

144, u5,0 0

By Bender-schmidt realtion,

ui, j 11

[u2

i 1, j ui 1, j ]

The values of u upto 4 sec are tabulated as follows

j\i 0 1 2 3 4 5

00.511.522.533.54

000000000

244242393026.62519.87517.531213.0312

8484786053.2539.7535.062526.062522.9687

1441447867.549.543.532.2528.406221.0938

14472573933.7524.7521.7516.12514.2031

000000000

x

a

a a

2


143

CIVIL ENGINEERING

ui

01, j

u0

u

0 ju4

u

x

5.4 Solution of One dimensional wave equation

Introduction

The one dimensional wave equation is of hyperbolic type. In this chapter, we discuss the2

finite difference solution of the one dimensional wave equation utt a u .

Explicit method to solve utt xx

ui, j 1 2(1 2 2i, j

2 2i 1, j ui 1, j ui, j 1 (1)

Where =k/hFormula (1) is the explicit scheme for solving the wave equation.

Problems1.Solve numerically , 4uxx utt with the boundary conditions u(0,t)=0,u(4,t)=0 and the initialconditions ut (x,0)Sol:

Here a2=4

A=2 and h=1

0 & u(x,0) x(4 x), taking h=1.Compute u upto t=3sec.

We choose k=h/a k=1/2

The finite difference scheme is

ui, j 1 i 1, j i, j 1

u(0, t), j & u(4, t) 0 , j

u(x,0) x(4 ) ui,0 i(4 i),i 0,1,2,3,4

u0,0 0, u1,0

3, u2,0

3, u4,0 0

u1,1 4 0 / 2 2u2,1 3, u3,1 2

The values of u for steps t=1,1.5,2,2.5,3 are calculated using (1) and tabulated below.

j\i 0 1 2 3 40 0 3 4 3 01 0 2 3 2 02 0 0 0 0 03 0 -2 3 -2 04 0 -3 -4 3 0


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5 0 -2 -3 -2 06 0 0 0 0 0



x(4x)

ui

01, j

u0

u

0 ju4

u

x

1

1

1

1

1

2 ut 2

2 ux 2

ut u

2.Solve uxx utt given u(0,t)=0,u(4,t)=0,u(x,0)= u(x,0) & ut (x,0)2

0. Take h=1.Find

the solution upto 5 steps in t-direction.

Sol:

Here a2=4

A=2 and h=1

We choose k=h/a k=1/2

The finite difference scheme is

ui, j 1 i 1, j i, j 1

u(0, t), j & u(4, t) 0 , j

u(x,0) x(4 ) / 2

ui,0 i(4 i) / 2, i

0,1,2,3,4

u0,0

u1,1

u2,1

0, u1,0

.5, u3,1

.5, u2,0

2, u3,0

.5, u4,0 0

The values of u upto t=5 are tabulated below.

j\i 0 1 2 3 40 0 1.5 2 1.5 01 0 1 1.5 1 02 0 0 0 0 03 0 -1 -1.5 -1 04 0 -1.5 -2 -1.5 05 0 -1 -1.5 -1 0

Exercise

1.Solve ,0<x<1,t>0 given that u(x,0)=100sinπx, (x,0) 0, u(0, t)

(1, t) 0

Taking h=1/4 compute u for 4 time steps.


MA6459 NUMERICAL METHODS2 ut 2

2 ux 2 uu

t

2. Solve ,0<x<1,t>0 given that u(x,0)=100(x-x2), (x,0) 0, u(0, t)

(1, t) 0

compute u for 4 time steps taking h=0.25.



1

Tutorial problemsTutorial-1

1. Solve the boundary value problem ymethod.

Ans:0.129,0.147

64y 10 0 with y(0)=y(1)=0 by the finite difference

2. Determine the values of y at the pivotal pointa of the interval (0,1) if y satisfies the boundaryvalue problem

yiv 81y 81x2 , y(0) y(1) y (0) y (1) 0 take n=3.

Ans:0.1222,0.1556

3. Write down the finite difference analogue of the equation uxx uyy 0 solve it for the regionbounded by the squre 0 x 4,0 y 4 and the boundary conditions being given as,u=0 at

x=0;u=8+2y at x=4 ; u x2 / 2 at y=0; u

x2 at y=4.with h=k=1,use Gauss-seidel method to

compute the values at the initial at the internal mesh points.

Ans:2,4.9,9,2.1,4.7,8.1,1.6,3.7,6.6

4. Solve the poisson equation uxx

uyy 81xy,0 x 1;0y

1and

u(0,y)=u(x,0)=0,u(x,1)=u(1,y)=100 with the square meshs,each of length h=1/3

Ans:51.08,76.54,25.79

5. Solve by Crank- Nicholson’s implicit method, ut

uxx ,0 x

1,t

0 with

u(x,0) 100 (x

x2 ), u(0,t) 0,u(1,t) 1. compute u for one time step with h=0.25.

Ans:9.82,14.29

Tutorial -2

2. Solve the boundary value problem ymethod.

64y 10 0 with y(0)=y(1)=0by the finite difference

Ans: y(0.5)=0.147,y(0.25)=y(0.75)=0.129

3. Solve xy

y 0, y (1)

0, y(2)

with h=0.5



Ans: y (0)=1.6552 and y(0.5)=1.4483

4. When steady condition prevail,the temperature distribution of the plate is representd by Laplceequation uxx uyy 0 .The temperature along the edges of the square plate of side 4 are given by

u=0 along x=y=0;u=x3 along y=4 and u=16y alon x=4.Divide the square plate into 16 square

2



23

x 2 j

meshes of side h=1,compute the temperatures at all of the 9 interior grid points by Leibmann’s iteration process.

Ans: u1 4.5, u2

12.5, u3

26.9, u4

4.4, u5

10.7, u6

20.0, u7

2.6, u8

6, u9

10.5

5. Solve the Poisson’s equation

2u 8x 2 y 2

inside a square region bounded by the lines

x 2, y 2,u 0 on the boundary .Assume the origin at the centre of the square and divide the

square into 16 equal parts.

Ans:u1 u3 u7 u9

u2 u4 u6 u8 u5

6. Solve the Laplace equation 2u 0 inside the square region bounded by the linesx=0,x=4,y=0,and y=4 given that u

x 2 y 2

on the boundary..

Ans: u1 22, u2

55.5, u3

99.7, u4

16.6, u5

36, u6

55, u7

8.3, u8

16.6, u9 22

Tutorial -3

1.Find the values of the function u(x,y) satisfying the differential equation ut 4uxx and the

boundary conditions u(0,t)

,j=0,1,2,3,4,5.0 u(8,t) and u(x,0) 4x for points x=0,1,2,3,4,5,6,7,8, t

2 2

u2. Solve by the Crank-Nicholson’s method,1 u

,0

x 1, t

0, given that

t 16 x 2

u(x,0) 100sin x,u(0,t) u(1,t) 0.Compute u for one time step, taking h=1/4.

Ans: u(1

,1)4

38.67;u(1

,1)2

54.69, u( 3

,1)4

38.67



ut

2ux 2

3. Using Crank-Nicholson’s method solve

taking h=1/4 and k=1/8 ,subject to u(x,0) 0,u(0,t) 0,u(1,t) 16t.

Ans: u(1

, 1

) 0.0952;u(1

, 1

) 0.2856,u( 3

, 1

)0.7619.

4 8 2 8 4 8

4.339,5.341



Question Bank

Part A

1. What is the error for solving Laplace and Poisson’s equations by finite difference method?

Sol:

The error in replacing by the difference expression is of the order . Since h=k, the

error in replaing by the difference expression is of the order .

2. Define a difference quotient.Sol:

A difference quotient is the quotient obtained by dividing the difference between two values of a function by the difference between two corresponding values of the independent variable.

3. Why is Crank Nicholson’s scheme called an implicit scheme?Sol:

The Schematic representation of crank Nicholson method is shown below.The solution value at any point (i,j+1) on the (j +1)th level is dependent on the solution values at the neighboring points on the same level and on three values on the j th level. Hence it is an implicit method.

4. What are the methods to solve second order boundary-value problems?Sol:

(i)Finite difference method (ii)Shooting method.

5. What is the classification of one dimensional heat flow equation.Sol:

One dimensional heat flow equation isHere A=1,B=0,C=0

B2 − 4AC = 0Hence the one dimensional heat flow equation is parabolic.

6. State Schmidt’s explicit formula for solving heat flow equation

Sol:



+(1-2 ) if , = .



7. Write an explicit formula to solve numerically the heat equation (parabolic equation)

Sol:

+(1-2 ) Where (h is the space for the variable x and k is the space in the time direction).The above formula is a relation between the function values at the two levels j+1 and j and iscalled a two level formula. The solution value at any point (i,j+1) on the (j+1)th level is expressed in terms of the solution values at the points (i-1,j),(i,j) and (i+1,j) on the j th level.Sucha method is called explicit formula. the formula is geometrically represented below.

8. State the condition for the equation to be(i) elliptic,(ii)parabolic(iii)hyperbolic when A,B,C are functions of x and ySol:

The equation is elliptic if (2B2 ) − 4AC < 0(i.e) B2 − AC < 0. It is parabolic if B2 − AC = 0 and hyperbolic if B2− 4AC > 0

9. Write a note on the stability and convergence of the solution of the difference equation corresponding to the hyperbolic equation .Sol:

For ,λ = the solution of the difference equation is stable and coincides with the

solution of the differential equation. For λ > ,the solution is unstable.

For λ < ,the solution is stable but not convergent.

10. State the explicit scheme formula for the solution of the wave equation.Sol:

The formula to solve numerically the wave equation =0 is

The schematic representation is shown below.The solution value at any point (i,j+1) on the ( j +1)th level is expressed in terms of solution values on the previous j and (j-1) levels (and not interms of values on the same level).Hence this is an explicit difference formula.

Part B

1.Reduce the following elliptic partial differential equations using orthogonal collocation in both the x and y directions:

+ =f(x,y);0 x,y 1, X=0; T= ; X=1; T= ; y=0; T= ; y=1; =0.



2. Solve = , subject to u(0,t)=u(1,t)=0 and u(x,0)=sin , 0<x<1, using Bender-Schmidt

method.

3.Obtain the finite difference from the first order partial differential equation

+ =0: - <x < , T(0,x)=T*(x).

4. Solve U xx = U tt with boundary condition u(0,t) = u(4,t) and the initial condition u t(x,0) = 0 , u(x,0)=x(4-x) taking h =1, k = ½ (solve one period).

5. Solve y tt = 4y xx subject to the condition y(0,t) =0, y(2,t)=0, y(x,0) = x(2-x),

(x, 0)= 0. Do 4steps and find the values upto 2 decimal accuracy.

6.By iteration method solve the elliptic equation uxx+uyy =0 over the square region of side 4,satisfying the boundary conditions.

(i) u(0,y)=0,0 y 4

(ii) u(4,y)=8+2y, 0 y 4

(iii) u(x,0)= , 0 x 4

(iv) u(x,4)=x2, 0 x 4.

7.Using Crank-Nicolson’s scheme, solve 16 = , 0 x 1,t>0 subject to u(x,0)=0, u(0,t)=0,

u(1,t)=100t. Compare u for one step in t direction taking h=1/4.

8.Solve utt=uxx; 0<x<2, t>0 subject to u(x,0)=0,ut(x,0)=100(2x-x2), u(0,t)=0,u(2,t)=0,choosing h=1/2 compute u for four time steps.

9.Solve by Bender Schmidt formula upto t=5 for the equation uxx=ut, subject to

u(0,t)=0, u(5,t)=0 and u(x,0)=x2(25-x2), taking h=1.

10. Find the pivotal values of the equation = with given conditions

u(0,t)=0=u(4,t), u(x,0)=xu(4-x) and by taking h=1 for 4 time steps.

11. Obtain the Crank-Nicholson finite difference method by taking = 1. Hence find

u(x ,t ) in the rod for two time steps for the heat equation , given

u (x ,0) =sin , u(0, t) =0, u(1,t, )= 0. Taking h= 0.2.



12. .Solve 8x2y2 for square mesh given u= 0 on the four

boundaries dividing the square into 16 sub-squares of length 1 unit.



(ii) Solve for a positive root of x – cos x = 0 by Regula Falsi Method. (6)

ma6459numerical methodsl t p c 3 1 0 4 - erode ... · web viewunit vboundary value problems in...

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